Astronomy

Calculating the position angle of the moon (the rolling back and forth)

Calculating the position angle of the moon (the rolling back and forth)


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I have been watching the the NASA Moon Phases Video of the year 2020. Pretty much everything makes sense and I know how to calculate it including the moon phases, declination, azimuth, apogee and perigee except one thing, the position angle of the moon. Its donated by Pos. Angle at the bottom right of that video.

Apparently when the Pos. angle is 0, that means when the azimuth of the moon reaches 180 looking from the northern hemisphere, the phase's north direction of the moon is going to be perfectly aligned with the north direction on earth(180 azimuth):

Given the moon and sun current declination, How do you calculated the Position angle of the phase as it's described in NASA Moon phase 2020 ? I'm looking for an output that matches NASA's calculations or close to it at any given date. Any help would be much appreciate…


File:Phase and libration of the Moon at hourly intervals (2012).ogv

This animation shows the geocentric phase, libration, position angle of the axis, and apparent diameter of the Moon throughout the year 2012, at hourly intervals.

The jagged, cratered, airless lunar terrain casts sharp shadows that clearly outline the Moon's surface features for observers on Earth. This is especially true near the terminator, the line between day and night, where surface features appear in high relief. Elevation measurements by the Lunar Orbiter Laser Altimeter (LOLA) aboard the Lunar Reconnaissance Orbiter (LRO) make it possible to simulate shadows on the Moon's surface with unprecedented accuracy and detail.

The Moon always keeps the same face to us, but not exactly the same face. Because of the tilt and shape of its orbit, we see the Moon from slightly different angles over the course of a month. When a month is compressed into 12 seconds, as it is in this animation, our changing view of the Moon makes it look like it's wobbling. This wobble is called libration.

The word comes from the Latin for "balance scale" (as does the name of the zodiac constellation Libra) and refers to the way such a scale tips up and down on alternating sides. The sub-Earth point gives the amount of libration in longitude and latitude. The sub-Earth point is also the apparent center of the Moon's disk and the location on the Moon where the Earth is directly overhead.

The Moon is subject to other motions as well. It appears to roll back and forth around the sub-Earth point. The roll angle is given by the position angle of the axis, which is the angle of the Moon's north pole relative to celestial north. The Moon also approaches and recedes from us, appearing to grow and shrink. The two extremes, called perigee (near) and apogee (far), differ by more than 10%.

The most noticed monthly variation in the Moon's appearance is the cycle of phases, caused by the changing angle of the Sun as the Moon orbits the Earth. The cycle begins with the waxing (growing) crescent Moon visible in the west just after sunset. By first quarter, the Moon is high in the sky at sunset and sets around midnight. The full Moon rises at sunset and is high in the sky at midnight. The third quarter Moon is often surprisingly conspicuous in the daylit western sky long after sunrise.


Sun and Moon Anomaly

  “  Below is a photo that my son took in Scotland showing the sun and moon at the same time. I immediately noticed this anomaly that the light illuminating the moon could not possibly come from the sun. I sent the photo to 4 University astronomy departments and only one responded and that was Cambridge University which is near where I live. The response came from the department librarian (not an astronomer) who said he had never heard of this before. He gave me two possible solutions, one was was from an engineer (not an astronomer) in which he got confused between perspective and light ray tracing and the other was referring to Einstein's theory of light bending by gravity. I check out Einstein and the effect was so small as to be almost immeasurable.  ”

  “  I have looked at the various 'complex' explanations for what to me is a very simple model. What need is there to introduce 'curved planes' and 'starry sky domes' all of which do not exist in reality? It is only referred to as an 'illusion' because observation doesn't fit the conventional model hence the complex explanations to try and make it work. The anomaly is acknowledged to exist with or without photos. Since everyone believes that the moon is illuminated by the sun then simple normal physics do not seem to work. Either the physics is wrong or the sun does not illuminate the moon. I realise that is a heavy statement!

Therefore I state once again:

  1. The sun and the moon are two objects (like a torch and a football) that are suspended in a 3 dimensional space and size should not matter.
  2. The moon/football are illuminated by the sun/torch and a perpendicular line or light ray can be drawn between them.
  3. It doesn't matter where in space you choose to view them, a perpendicular line or light ray can still be drawn between them.

This drawing explains my doubts:  ”

  “  I'm very surprised that some of you have never noticed it before hence the suggestion asking me to post a video. This is a very common occurrence and I have seen it many many times as I go for my morning walk at about 8.00am every morning. I have never thought of actually tabulating my observations.  ”


How to Make a Clinometer

This article was co-authored by Bess Ruff, MA. Bess Ruff is a Geography PhD student at Florida State University. She received her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group.

There are 7 references cited in this article, which can be found at the bottom of the page.

wikiHow marks an article as reader-approved once it receives enough positive feedback. This article received 28 testimonials and 83% of readers who voted found it helpful, earning it our reader-approved status.

This article has been viewed 487,006 times.

A clinometer, also called a declinometer or an inclinometer, is an instrument that measures vertical slope, usually the angle between the ground or the observer and a tall object. A simple, or fixed angle, clinometer requires plenty of room to walk back and forth when measuring an object. A protractor clinometer lets you measure while standing in place, and is an easy-to-make version of the clinometers frequently used in astronomy, surveying, engineering, and forestry.


3.E: Energy (Exercises)

  • Contributed by Timon Idema
  • Associate Professor (Bionanoscience) at Delft University of Technology
  • Sourced from TU Delft Open
  1. Show that, if you ignore drag, a projectile fired at an initial velocity (v_0) and angle ( heta) has a range R given by
  2. A target is situated 1.5 km away from a cannon across a flat field. Will the target be hit if the firing angle is (42^) and the cannonball is fired at an initial velocity of 121 m/s? (Cannonballs, as you know,do not bounce).
  3. To increase the cannon&rsquos range, you put it on a tower of height (h_0). Find the maximum range in this case, as a function of the firing angle and velocity, assuming the land around is still flat.

3.2 You push a box of mass m up a slope with angle ( heta) and kinetic friction coefficient (mu). Find the minimum initial speed v you must give the box so that it reaches a height h.

3.3 A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is (mu _1), and in region 2, the coefficient is (mu _2). Our objective is to find the net work W done by friction in pulling the board directly from region 1 to region 2, under the assumption that the board moves at constant velocity.

  1. Suppose that at some point during the process, the right edge of the board is a distance x from the boundary, as shown. When the board is at this position, what is the magnitude of the force of friction acting on the board, assuming that it&rsquos moving to the right? Express your answer in terms of all relevant variables (L, M, g, x, (mu _1), and (mu _2)).
  2. As we&rsquove seen in Section 3.1, when the force is not constant, you can determine the work by integrating the force over the displacement, (W= int F(x) dx). Integrate your answer from (a) to get the net work you need to do to pull the board from region 1 to region 2.

3.4 The government wishes to secure votes from car-owners by increasing the speed limit on the highway from 120 to 140 km/h. The opposition points out that this is both more dangerous and will cause more pollution. Lobbyists from the car industry tell the government not to worry: the drag coefficients of the cars have gone down significantly and their construction is a lot more solid than in the time that the 120 km/h speed limit was set.

  1. Suppose the 120 km/h limit was set with a Volkswagen Beetle ((=0.48)) in mind, and the lobbyist&rsquoscar has a drag coefficient of 0.19. Will the new car need to do more or less work to maintain a constant speed of 140 km/h than the Beetle at 120 km/h?
  2. What is the ratio of the total kinetic energy released in a full head-on collision (resulting in an immediate standstill) between two cars both at 140 km/h and two cars both at 120 km/h?
  3. The government dismisses the opposition&rsquos objections on safety by stating that on the highway, all cars move in the same direction (opposite direction lanes are well separated), so if they all move at 140 km/h, it would be just as safe as all at 120 km/h. The opposition then points out that running a Beetle (those are still around) at 120 km/h is already challenging, so there would be speed differences between newer and older cars. The government claims that the 20 km/h difference won&rsquot matter, as clearly even a Beetle can survive a 20 km/h collision. Explain why their argument is invalid.

3.5 Nuclear fusion, the process that powers the Sun, occurs when two low-mass atomic nuclei fuse together to make a larger nucleus, releasing substantial energy. Fusion is hard to achieve because atomic nuclei carry positive electric charge, and their electrical repulsion makes it difficult to get them close enough for the short-range nuclear force to bind them into a single nucleus. The figure below shows the potential-energy curve for fusion of two deuterons (heavy hydrogen nuclei, consisting of a proton and a neutron).The energy is measured in million electron volts ((MeV, 1 eV=1.6 cdot 10^ <-19>J)), a unit commonly used in nuclear physics, and the separation is in femtometers ((1 fm=10^ <-15>m)).

  1. Find the position(s) (if any) at which the force between two deuterons is zero.
  2. Find the kinetic energy two initially widely separated deuterons need to have to get close enough to fuse.
  3. The energy available in fusion is the energy difference between that of widely separated deuterons and the bound deutrons after they&rsquove &lsquofallen&rsquo into the deep potential well shown in the figure. About how big is that energy?
  4. Determine whether the force between two deuterons that are 4 fm apart is repulsive, attractive, or zero.

3.6 A pigeon in flight experiences a drag force due to air resistance given approximately by (F=bv^2), where v is the flight speed and b is a constant.

  1. What are the units of b?
  2. What is the largest possible speed of the pigeon if its maximum power output is P?
  3. By what factor does the largest possible speed increase if the maximum power output is doubled
  1. For which value(s) of the parameters (alpha, eta, ext < and >gamma) is the force given by [oldsymbol=left(x^ <3>y^<3>+alpha z^<2>, eta x^ <4>y^<2>, gamma x z ight)] conservative?
  2. Find the force for the potential energy given by (U(x,y,z)=frac-frac).

3.8 A point mass is connected to two opposite walls by two springs, as shown in the figure. The distance between the walls is 2L. The left spring has rest length (l_1=frac<2>) and spring constant (k_1=k), the right spring has rest length (l_2=frac<3L><4>) and spring constant (k_2=3k).

  1. Determine the magnitude of the force acting on the point mass if it is at x=0.
  2. Determine the equilibrium position of the point mass.
  3. Find the potential energy of the point mass as a function of x. Use the equilibrium point from (b) as your point of reference.
  4. If the point mass is displaced a small distance from its equilibrium position and then released, it will oscillate. By comparing the equation of the net force on the mass in this system with a simple harmonic oscillator, determine the frequency of that oscillation. (We&rsquoll return to systems oscillating about the minimum of a potential energy in Section 8.1.4, feel free to take a sneak peak ahead).

3.9 A block of mass m=3.50 kg slides from rest a distance d down a frictionless incline at angle ( heta=30.0^circ),where it runs into a spring of spring constant 450 N/m. When the block momentarily stops, it has com-pressed the spring by 25.0 cm.

  1. Find d.
  2. What is the distance between the first block-spring contact and the point at which the block&rsquos speed is greatest?

3.10 Playground slides frequently have sections of varying slope: steeper ones to pick up speed, less steep ones to lose speed, so kids (and students) arrive at the bottom safely. We consider a slide with two steep sections (angle (alpha)) and two less steep ones (angle (eta)). Each of the sections has a width L. The slide has a coefficient of kinetic friction (mu).

  1. Kids start at the top of the slide with velocity zero. Calculate the velocity of a kid of mass m at the end of the first steep section.
  2. Now calculate the velocity of the kid at the bottom of the entire slide.
  3. If L=1.0 m, (alpha=30^circ) and (mu=0.5), find the minimum value (eta) must have so that kids up to 30 kg can enjoy the slide (Hint: what is the minimum requirement for the slide to be functional)?
  4. A given slide has (alpha=30^circ), (eta=20^circ), and (mu=0.5). A young child of 10 kg slides down, while its cousin of 20 kg sits at the bottom. When the sliding kid reaches the end, the two children collide, and together slide further over the ground. The coefficient of kinetic friction with the ground is 0.70. How far do the two children slide before they come to a full stop?

3.11 In this problem, we consider the anharmonic potential given b

where a, b, and (x_0) are positive constants.

  1. Find the dimensions of a, b, and (x_0).
  2. Determine whether the force on a particle at a position (x gt gt x_0) is attractive or repulsive (taking the origin as your point of reference).
  3. Find the equilibrium point(s) (if any) of this potential, and determine their stability.
  4. For b=0, the potential given in Equation (3.24) becomes harmonic (i.e., the potential of a harmonic oscillator), in which case a particle that is initially located at a non-equilibrium point will oscillate. Are there initial values for x for which a particle in this anharmonic potential will oscillate? If so,find them,and find the approximate oscillation frequency if not, explain why not. (NB: As the problem involves a third order polynomial function, you may find yourself having to solve a third order problem. When that happens, for your answer you can simply say: the solution x to the problem X).

3.12 After you have successfully finished your mechanics course, you decide to launch the book into an orbit around the Earth. However, the teacher is not convinced that you do not need it anymore and asks the following question: What is the ratio between the kinetic energy and the potential energy of the book in its orbit?

Let m be the mass of the book, (M_ ext < and >R_) the mass and the radius of the Earth respectively. The gravitational pull at distance r from the center is given by Newton&rsquos law of gravitation (Equation 2.2.3):

  1. Find the orbital velocity v of an object at height h above the surface of the Earth.
  2. Express the work required to get the book at height h.
  3. Calculate the ratio between the kinetic and the potential energy of the book in its orbit.
  4. What requires more work, getting the book to the International Space Station (orbiting at h=400 km)or giving it the same speed as the ISS?

3.13 Using dimensional arguments, in Problem 1.4 we found the scaling relation of the escape velocity (the minimal initial velocity an object must have to escape the gravitational pull of the planet/moon/other object it&rsquos on completely) with the mass of the radius of the planet. Here, we&rsquoll re-derive the result, including the numerical factor that dimensional arguments cannot give us.

  1. Derive the expression of the gravitational potential energy,Ug, of an object of mass m due to a gravitational force (F_g) given by Newton&rsquos law of gravitation (Equation 2.2.3) [F_>=-frac> hat] Set the value of the integration constant by ( ightarrow 0 ext < as >r ightarrow infty)
  2. Find the escape velocity on the surface of a planet of mass M and radius R by equating the initial kinetic energy of your object (when launched from the surface of the planet) to the total gravitational potential energy it has there.

3.14 A cannonball is fired upwards from the surface of the Earth with just enough speed such that it reaches the Moon. Find the speed of the cannonball as it crashes on the Moon&rsquos surface, taking the gravity of both the Earth and the Moon into account. Table B.3 contains the necessary astronomical data.

3.15 The draw force F(x) of a Turkish bow as a function of the bowstring displacement x (for x gt 0) is approximately given by a quadrant of the ellipse [left(frac> ight)^<2>+left(frac ight)^<2>=1] In rest, the bowstring is at x=0 when pulled all the way back, it&rsquos at x=-d.

  1. Calculate the work done by the bow in accelerating an arrow of mass m=37 g, for d=0.85 m, and Fmax=360 N.
  2. Assuming that all of the work is converted to kinetic energy of the arrow, find the maximum distance the arrow can fly.Hint: which variable can you control when shooting? Maximize the distance with respect to that variable.
  3. Compare the result of (b) with the range of a bow that acts like a simple (Hookean) spring with the same values of Fmax and d. How much further does the arrow shot from the Turkish bow fly than that of the simple spring bow?

3.16 A massive cylinder with mass M and radius R is connected to a wall by a spring at its center (see figure).The cylinder can roll back-and-forth without slipping.

  1. Determine the total energy of the system consisting of the cylinder and the spring.
  2. Differentiate the energy of problem (16a) to obtain the equation of motion of the cylinder and spring system.
  3. Find the oscillation frequency of the cylinder by comparing the equation of motion at (16b) with that of a simple harmonic oscillator (a mass-spring system).

3.17 A small particle (blue dot) is placed atop the center of a hemispherical mount of ice of radius R (see figure). It slides down the side of the mount with negligible initial speed. Assuming no friction between the ice and the particle, find the height at which the particle loses contact with the ice.

Hint: To solve this problem, first draw a free body diagram, and combine what you know of energy and forces.

3.18 Pulling membrane tubes

The (potential) energy of a cylindrical membrane tube of length L and radius R is given by

Here (kappa) is the membrane&rsquos bending modulus and (sigma) its surface tension.


The Physics Of Bowling

A bowling ball is made from urethane, plastic, reactive resin or a combination of these materials. Ten-pin bowling balls generally have three holes drilled into them two finger holes and one thumb hole for gripping. For ten-pin bowling, regulating bodies allow for a maximum weight of 16 lb (7.2 kg), and a maximum diameter of 8.6 inches (21.8 cm) (ref: http://en.wikipedia.org/wiki/Bowling_ball).

The physics of bowling discussed here will be with regards to ten-pin bowling, which is one of the most common sports in the game of bowling.

The figure below shows two ten-pin balls.

Source: http://en.wikipedia.org/wiki/Bowling_ball. Author: http://en.wikipedia.org/wiki/User:Ommnomnomgulp

Bowling Ball Interior

The bowling ball consists of a hard outer shell with a weight block in the core (the inside of the bowling ball). The mass and shape of the weight block affects the spin of the bowling ball and how it curves as it rolls down the lane. These play an important role in the physics of bowling and (consequently) a bowler's performance, as will be discussed.

There are two basic types of weight blocks used, symmetric weight blocks and asymmetric weight blocks. To illustrate them, imagine cutting a bowling ball in half with an imaginary cutting plane (as shown below) so as to expose the full cross-section of the weight block inside the bowling ball.

The figure below shows the cross-section view for a symmetric weight block.

The weight block is molded into the bowling ball. The "pin" represents the topmost position of the weight block, as shown. This topmost position is closest to the outside surface of the bowling ball.

A symmetric weight block is called "symmetric" because it is axi-symmetric, meaning it has an axis of symmetry along its centerline. To illustrate, consider the figures below with a coordinate system xyz as shown. Note that point G represents the center of mass of the bowling ball.

Let the plane x-z represent the imaginary cutting plane. For any angle θ the resulting cross-section view of the symmetric weight block would be the same. In other words, a symmetric weight block is axi-symmetric with respect to angle θ.


The figure below shows the cross-section view for an asymmetric weight block.

Just like the symmetric weight block, the "pin" represents the top position of the asymmetric weight block. However, an asymmetric weight block is not symmetric with respect to angle θ. And due to its non-symmetry, a "PSA indicator pin" is placed on the side of the weight block, at the point (or area) closest to the outside surface of the bowling ball. These two pins are located 90° from each other.

The "PSA indicator pin" is also called the "mass bias" location. This is simply a different naming convention in bowling terminology.

The pins allow one to determine the orientation of the weight block inside the bowling ball. A symmetric weight block only needs a single pin to define its orientation inside the bowling ball, but an asymmetric weight block needs two pins to define its orientation (due to its non-symmetry).

The pins give important information on where to drill the finger and thumb holes of the bowling ball so that the bowler can control the spin and curve of the ball, in order to make the best possible shot. This will be explained in more detail later on, as the physics of bowling is discussed in greater depth.


Bowling balls with symmetric and asymmetric weight blocks result in similar performance, but bowling balls with an asymmetric weight block allow for a bit more "tweaking" to get the ball to react a certain way when in motion. However, the physics of bowling is very similar whether the weight block is symmetric or asymmetric.


In addition to the pins, two other points of interest on a bowling ball are the Positive Axis Point (PAP) and the CG point. These directly relate to the physics of bowling and are shown in the figure below.

The PAP is the initial axis of rotation of the bowling ball as soon as it begins traveling down the lane. The orientation of this axis depends entirely on the bowler's release technique. It is unique to each bowler.

wo is the initial rotation speed (angular velocity) of the bowling ball. This also depends on the bowler's technique.

The figure below shows an imaginary line originating from the geometric center C of the bowling ball, and passing through the center of mass G of the bowling ball. The intersection of this line with the surface of the bowling ball is called CG. This point is useful because it gives information on the location of the center of mass G of the bowling ball.

Note that the distance between C and G is greatly exaggerated for clarity. In reality, the distance between the two is very small and is typically less than 1 mm (ref: What Makes Bowling Balls Hook?, Cliff Frohlich, American Association of Physics Teachers, 2004).

The optimal trajectory of a bowling ball is a curved path where it strikes the pins at an angle. Striking the pins at an angle improves the chances that there will be a "strike" in which all the pins are knocked down.

Analyzing the physics of bowling is very useful because it allows one to understand the factors that influence how the bowling ball curves, and how one can make the best possible shot when striking the pins.

If the ball follows a curved path, it will be able to strike the pins at a greater angle than a bowling ball that travels in a straight line. Therefore, controlling the degree of curve is essential to making the best possible shot.

For ten-pin bowling the length of the lane is 18 m (60 ft). The lane is oiled to protect it from wear, especially during the initial sliding (skidding) stage of the bowling ball, before it begins pure rolling.

The bowler must make the shot behind the foul line and must avoid getting the ball into the gutter. The two figures below illustrate the motion of a bowling ball as it travels down the lane. The ball motion shown is typical for a right-handed bowler. For a left-handed bowler the motion is simply "mirrored" so that the ball curves towards the left gutter instead of the right gutter, before hitting the pins.

The ball begins rolling with a rotation speed (angular velocity) of wo and a linear velocity Vo.

Typically, during the first part of the motion the bowling ball slides along the lane, since its rotational speed does not match with the linear velocity of the ball. But eventually, lane friction stops the ball from sliding, and pure rolling begins. The ball then continues rolling until it hits the pins.

The ball hits the pins at an angle θ. The greater this angle, the more oblique the impact and the greater the chance that all the pins will be knocked down. The deflection δ is called "hook". It is the sideways deflection of the ball from the original (straight) trajectory indicated by the dashed blue line.

Ideally, the front-most pins are hit first by the bowling ball (at an oblique angle), since this will most likely result in a strike.

In order to model the physics of bowling, a (fixed) coordinate axes XYZ is chosen with the following orientation: The X-axis is aligned with the right gutter (parallel to the lane), the Y-axis is aligned with the foul line, and the Z-axis is vertical (perpendicular to the lane). This chosen coordinate system becomes relevant in the next section, where the physics is analyzed in depth.


The location of wo (PAP) relative to the pin location(s) on the bowling ball, is very important in influencing the amount of hook δ, and impact angle θ. Therefore, the finger and thumb holes must be positioned in an optimal way relative to the pin location(s), so as to optimize δ and θ, in order to get the best possible shot.

The location of CG relative to the PAP location also influences δ and θ, but to a much lesser degree, so it is not accounted for nearly as much when optimizing ball performance.

For each individual bowler, the PAP is (approximately) in a fixed location relative to the finger and thumb holes. For each individual bowler the position of the PAP relative to the finger and thumb holes, must be determined with a "test" ball. And once this relative position is determined, the finger and thumb holes are drilled in a new ball, so that the resulting PAP is in an optimal position relative to the pin location(s).

In a nutshell, the location of the PAP relative to the pins on the bowling ball determines how much the bowling ball precesses as it travels down the lane. Consequently, the level of precession is directly proportional to the level of friction between the lane and the bowling ball. The level of friction, in turn, has a large influence on δ and θ.

So in general, more precession leads to more friction and results in more hook, and less precession leads to less friction and results in less hook. This is one of the main criterion that all good players are aware of when calibrating their game for best performance.

The figure below illustrates precession, and how it affects the level of friction between the lane and the bowling ball.

Precession is the change in direction of the bowling balls spin vector wspin, over time. For example, a spinning top (pivoted about a base) has two spin components. The first component is the main spin, which is the spinning of the top about its central axis. The second component is the secondary spin, which causes the orientation of the top to change so that its central axis goes around in a circle. This secondary spin component is called precession.

In the case of the bowling ball, precession causes a back and forth rocking (as shown) as it travels down the lane. This changes the contact surface after each full ball revolution, and thus increases the area of the ball in contact with the lane. This becomes evident by the various lines of oil (shown in the figure above) that the ball picks up from contact with the lane. The oil lines appear “flared out”, and the width of the flare (called "track flare") is a measure of the degree of rocking (due to precession). Each line of oil represents one full revolution of the ball as it travels down the lane.

Therefore, precession results in greater friction between the lane and the bowling ball, since the oil is "spread out" on the surface of the bowling ball in a larger area than if the ball did not precess. If the bowling ball did not precess, there would only be one oil line, and this would result in lower friction due to repeated lane contact with the same area of the bowling ball. Hence, precession results in a greater area of the bowling ball making contact with the lane, which results in lower oil accumulation on the ball per contact area, thus resulting in greater friction. This becomes most apparent on the "dry" (non oiled) part of the lane, which is typically the last third (or so) of the lane length. In this part of the lane, friction between lane and ball is most sensitive to the amount of oil accumulated on the ball per contact area (since the lane itself is dry). This means that precession raises the level of friction between lane and ball. Therefore, in the dry part of the lane a precessing ball will hook the most.

The figure below shows another illustration of the typical oil line pattern one might see on a bowling ball that has undergone precession.


As mentioned already, the position of the PAP relative to the pin locations determines how much the bowling ball precesses as it travels down the lane. In particular, we can specify the relative position of the PAP for bowling balls with symmetric and asymmetric weight blocks. This is a key factor in the physics, since it directly affects the motion of the ball as it travels down the lane. It is described next.


Bowling Balls With Symmetric Weight Blocks

If one wishes to have no precession, the PAP must be located at the topmost pin location or anywhere on the circumference around the middle of the bowling ball, represented by the dotted line, shown in the figure below.

If one wishes to have precession, then the PAP must be located between these two locations - that is, between the topmost pin and the dotted line. Experienced ball drillers know where to place the PAP to get the desired amount of precession. To have maximum precession (maximum track flare) the PAP must be placed exactly in between the topmost pin and the dotted line, so that the distance from the PAP to the topmost pin is equal to the distance from the PAP to the dotted line.

In physics terms, an axis passing through G and the topmost pin location represents the minimum principal moment of inertia, and an axis passing through G and any point on the dotted line represents the maximum principal moment of inertia - this is due to the symmetry of the weight block.

Side note: Precession is a form of unstable rotation, and it occurs when the PAP does not coincide with the principal axes for the minimum or maximum moment of inertia.

For stable rotation (no precession), the PAP must coincide with the principal axes for the minimum or maximum moment of inertia.

Bowling Balls With Asymmetric Weight Blocks

If one wishes to have no precession, the PAP must be located at the topmost pin location or at the PSA indicator pin location on the side, as shown in the figure below.

If one wishes to have precession, then the PAP must be located between these two locations - that is, between the topmost pin and the PSA pin. Experienced ball drillers know where to place the PAP to get the desired amount of precession. To have maximum precession (maximum track flare) the PAP must be placed exactly in between the topmost pin and the PSA pin, so that the distance from the PAP to the topmost pin is equal to the distance from the PAP to the PSA pin.

In physics terms, an axis passing through G and the topmost pin location represents the minimum principal moment of inertia, and an axis passing through G and the PSA pin represents the maximum principal moment of inertia - this is due to the asymmetry of the weight block. Note that PSA means "preferred spin axis" since the axis with the maximum principal moment of inertia is naturally the most stable axis of rotation for a rigid body.


So far in discussing the physics of bowling, we have talked about the influence of PAP location and friction on ball motion. But there is another factor which influences ball motion, although not as much. It is the position of the CG. If the CG is on the left side of the bowling ball as it travels down the lane, the ball tends to hook a bit more (to the left), so it's beneficial. The figure below illustrates this.


Simulation For Different Friction Cases

The main influence on ball motion is friction between the ball and the lane, whether it's due to friction influenced by ball precession, or natural lane conditions (e.g. oiled vs. non-oiled).

Thus, we wish to look at three different cases of friction, in order to observe its effect on ball motion.

Let's assume that the initial angular velocity of the ball wo = 30 rad/s, and the initial linear velocity of the ball Vo = 8 m/s. This is a reasonable estimate for the typical bowler who averages 200 (ref: What Makes Bowling Balls Hook?, Cliff Frohlich, American Association of Physics Teachers, 2004).

Let's assume wo is pointing 15° to the left of the Y-axis, so that it has components wX = -30×sin 15° = -7.76 rad/s, and wY = 30×cos 15° = 28.98 rad/s.

And let's assume that Vo is pointing in the positive X-direction.

The beginning of the bowling balls trajectory is at X = 0.


The following additional input values are used with regards to the initial orientation, and properties of the bowling ball:

• The distance between the center of the bowling ball C and the center of mass G is 1 mm (0.001 m). The point CG is pointing in the positive Y-direction.

• The bowling ball has a symmetric weight block with the pin pointing in the positive Y-direction.

• The radius of the bowling ball is 10.85 cm.

• The mass of the bowling ball is 7 kg.

• The minimum principal moment of inertia is 0.031 kg·m 2 . This is the moment of inertia about an axis passing through the center of mass of the bowling ball G and the pin.

• The maximum principal moment of inertia is 0.033 kg·m 2 . This is the moment of inertia about any axis which passes through the center of mass of the bowling ball G and which is perpendicular to the axis corresponding to the minimum principal moment of inertia.

Lastly, the input value for the acceleration due to gravity (on earth's surface) is 9.8 m/s 2 .


The above input values will be held constant for the following three friction cases to be considered. In all three cases the static friction will be assumed sufficient enough to maintain pure rolling once pure rolling begins.

The coefficient of kinetic (sliding) friction is equal to 0.12. This value of kinetic friction will be taken as constant over the entire length of the lane, as illustrated below.

The bowling ball slides 9 m before pure rolling begins.

The ball hooks to the left δ = 68 cm.

The impact angle θ is 3.6°

The coefficient of kinetic friction is equal to 0.08. This value of kinetic friction will be taken as constant over the entire length of the lane, as illustrated below.

The bowling ball slides 13.7 m before pure rolling begins.

The ball hooks to the left δ = 55 cm.

The impact angle θ is 3.3°

The coefficient of kinetic friction is equal to 0.04 for the first 12 m of the lane, and is equal to 0.2 for the remainder of the lane (the remainder of the lane is "dry", no oil). This is illustrated below.

The bowling ball slides 15 m before pure rolling begins.

The ball hooks to the left δ = 37 cm.

The impact angle θ is 3.3°


Looking at the three friction cases we can see that the greater the level of friction, the greater the ball hook. The impact angle also increases slightly with friction.

As mentioned before, when the bowling ball precesses, the effective friction between the lane and ball increases. However, in the simulation, the friction coefficient was input as constant values (independent of ball precession), so the dependence of friction on ball precession was not captured in the model (due to the difficulty in doing so). So, to approximate the effect of friction, the friction coefficient was varied over a range of values in order to observe the effect on ball motion, and it is reasonable that this will approximate the influence of friction on ball motion when friction is affected by ball precession.

Hence, the simulation results indirectly show that precession causes the bowling ball to hook more (since precession increases friction).


A Closer Look At The Principal Moments Of Inertia

In general, a rigid body has three distinct principal moments of inertia. It is useful to know the magnitude and direction of these three quantities because they simplify the equations for three-dimensional rigid body dynamics .

The shape of the weight block inside a bowling ball makes it fairly straightforward to determine the principal directions of inertia. If a rigid body has two or three planes of symmetry, the principal directions will be aligned with these planes. By inspection, one can usually find the symmetry planes in the weight blocks, and as a result easily determine the principal directions for them. Since the remainder (outer part) of the bowling ball is itself almost completely symmetric due to its spherical shape, the result is that the (entire) bowling ball has principal directions closely aligned with the principal directions of the weight block.

For example, consider a bowling ball with a symmetric weight block, as shown in the two figures below.

The minimum principal moment of inertia of the bowling ball is about the x-direction. Let's call this Ix. This is the minimum moment of inertia because the mass of the weight block is the least "spread out" about the x-direction. This is a result of the design of the weight block.

The y and z directions correspond to the principal directions of the remaining two principal moments of inertia. These two moments of inertia are greater than Ix because the mass of the weight block is most "spread out" about these directions. However, because of symmetry, these two principal moments of inertia are in fact equal (Iy = Iz). Furthermore, due to symmetry these two moments of inertia are constant for all angles of θ.

Now, consider a bowling ball with an asymmetric weight block, as shown in the figure below.

Once more, the minimum principal moment of inertia of the bowling ball is given by Ix because the mass of the weight block is the least "spread out" about the x-direction.

The maximum principal moment of inertia is given by Iz because the mass of the weight block is most "spread out" about this direction. The remaining principal moment of inertia is given by Iy (pointing out of the page). Due to the non-symmetry of the weight block all three principal moments of inertia are distinct.

In physics, the moment of inertia (I) is sometimes defined as I = (mass)×(radius of gyration) 2 . So for a body of given mass, a higher radius of gyration results in a greater moment of inertia (by definition).

In bowling terminology, the radius of gyration is called the "RG value", and the term "RG differential" refers to the difference between the maximum and minimum moment of inertia of a bowling ball (i.e. ImaxImin).

In bowling lingo: The greater the "RG differential", the greater the potential for a bowling ball to create "track flare", and the greater the hook potential. Translated into physics lingo this means: The greater the difference ImaxImin, the greater the potential for precession. As a result, there is a greater potential for friction between ball and lane, and a greater potential for the ball to hook.

However, bowling regulations limit the amount of "RG differential" in a bowling ball.

Of course, whether or not this hook potential is fully utilized depends on where the PAP is placed relative to the pin locations on the bowling ball.


Physics Of Bowling – Model Description

I created a numerical model in Excel which captures the physics of bowling and simulates the motion of a bowling ball as it travels down the lane. A time step of 0.0001 seconds was used to ensure sufficient numerical accuracy. It was a challenging and time consuming task to develop this model, but the result is that I am able to accurately capture the essential physics behind bowling, without any gross simplifications and shortcuts which sacrifice accuracy.

To download the Excel spreadsheet right-click on this link.

I'm making the spreadsheet available, and the instructions to use it, for free. The download file is in compressed "zip" format. You need to uncompress this file before you can use it.

To use the bowling simulator you need to have Microsoft Excel installed on your computer. The program is compatible with all versions of Excel.


The development of the equations to fully analyze the physics behind bowling is quite an onerous task both to derive and fully present on a website. Therefore, the full mathematical development will not be presented here. Instead, the basic equations will be introduced in order to give the reader a basic understanding of the core physics and mathematics required.

The following assumptions are made in the model:

• The lane is perfectly flat.

• Friction is proportional to normal force, and the coefficient of friction between ball and lane is constant.

The figure below illustrates the full set up for analyzing the physics, with (fixed) global coordinate system XYZ as shown, along with sign convention.

C is the geometric center of the bowling ball

R is the radius of the bowling ball

G is the center of mass of the bowling ball

r is the vector from point C to point G. This vector is very short, usually less than 1 mm in length

P is the contact point between the bowling ball and the surface of the lane

mp is the vector from point G to point P

xyz is the local coordinate axes fixed to the bowling ball, so that it moves with the bowling ball. The orientation of xyz is such that it is aligned with the principal moments of inertia of the bowling ball.

w is the angular velocity of the bowling ball

α is the angular acceleration of the bowling ball

Vc is the linear velocity of the geometric center of the ball C. This is also considered to be the linear velocity of the bowling ball

ac is the linear acceleration of the geometric center of the ball C. This is also considered to be the linear acceleration of the bowling ball

Fs is the friction force acting on the bowling ball, due to contact between the ball and lane. This force is parallel to the lane surface

N is the normal force pushing up on the ball, perpendicular to the lane surface

g is the acceleration due to gravity, acting downwards. This value is equal to 9.8 m/s 2 on earth


Defining the orientation of xyz at any point in time

To define the orientation of xyz with respect to the (fixed) global XYZ axes, we must first set up the basic equations for the vectors x, y, and z in terms of direction cosines.

x, y, and z are unit vectors

l1, m1, n1 are the direction cosines corresponding to the unit vector x

l2, m2, n2 are the direction cosines corresponding to the unit vector y

l3, m3, n3 are the direction cosines corresponding to the unit vector z

I is the unit vector pointing along the positive direction of global X

J is the unit vector pointing along the positive direction of global Y

K is the unit vector pointing along the positive direction of global Z

By definition, the orientation of I, J, K remains constant, since XYZ is fixed in space.

Since x, y, and z are unit vectors,

and since x, y, and z are perpendicular to each other (by definition),


Defining the orientation of vector r at any point in time

Similar to before, the orientation of r is with respect to the (fixed) global XYZ axes:

rX is the component of r along the global X-direction

rY is the component of r along the global Y-direction

rZ is the component of r along the global Z-direction


Defining the orientation of angular velocity w and angular acceleration α at any point in time

Similar to before, the angular velocity and angular acceleration are vectors expressed relative to the (fixed) global XYZ axes. They are given by

α X is the component of α along the global X-direction

α Y is the component of α along the global Y-direction

α Z is the component of α along the global Z-direction

wX is the component of w along the global X-direction

wY is the component of w along the global Y-direction

wZ is the component of w along the global Z-direction

Thus, the angular velocity and angular acceleration of the bowling ball are always with respect to (fixed) ground.


Determining the components of the angular velocity w and angular acceleration α along the xyz axes

It's necessary to resolve the angular velocity and angular acceleration into their components along the xyz axis. This can be done using the vector dot product (ab):

where the subscripts x, y, and z (for the angular velocity and angular acceleration) denote their components along the x, y, and z directions, respectively.


Using forward integration to solve for the vectors w, r, and the orientation of xyz as a function of time

Let Δt be a small time step.

Forward integration is used to solve the physics equations. In this model, various unknown quantities at a new time t+Δt are calculated based on previous known quantities at time t.

For instance, we can solve for w(t+Δt) using the following equation for angular acceleration:

Similarly, we can find the new orientation of xyz at time t+Δt based on its previous orientation at time t:

where a×b represents the vector cross product.

Similarly, for the vector r:

Note that the above expressions for x, y, z, and r incrementally increase their vector lengths (magnitudes) for each time step. Therefore, we must normalize the vectors x, y, z, and r after each time step to preserve their length (magnitude). This normalizing operation can be expressed mathematically as follows:

where ro is the length of the r vector calculated from its original specification (at time t = 0).


The acceleration of the center of mass G

The center of mass G is offset from the geometric center of the bowling ball C. Therefore we must calculate the acceleration of G as follows:

aG is the acceleration of the center of mass G

aCX is the global X-component of the acceleration of point C

aCY is the global Y-component of the acceleration of point C


Application of Newton's Second Law

The external forces acting on the bowling ball are gravity and the contact forces at point P. Therefore, we can express the sum of the forces on the bowling ball as:

m is the mass of the bowling ball

FsX is the global X-component of the force acting on the bowling ball at point P

FsY is the global Y-component of the force acting on the bowling ball at point P

aGX is the global X-component of the acceleration of point G

aGY is the global Y-component of the acceleration of point G

aGZ is the global Z-component of the acceleration of point G


Sum of the moments (torque) acting on the bowling ball

If we take the sum of the moments about the center of mass G, the only forces to account for are those acting at the contact point P, with vector arm mp. (The force of gravity exerts no moment about point G since its line of action passes through point G).

We can thus apply the Euler equations of motion for a rigid body:

where Ix, Iy, and Iz are the principal moments of inertia, and

Note that the sum of moments in the Euler equations are expressed with respect to the local x, y, and z directions. It is necessary to do this when using these equations.


Lastly, we must consider two separate cases of ball motion when analyzing the physics. The first case involves kinetic (sliding) friction, where the bowling ball skids along the lane. The second case involves static friction, where the bowling ball has stopped skidding and is in a state of pure rolling.


Case 1 – Kinetic Friction

For this case, there is relative motion between the bowling ball and the lane at point P. We can express this relative motion as velocity VP, where:

VCX is the global X-component of the velocity of point C

VCY is the global Y-component of the velocity of point C

The magnitude of kinetic friction is given as

where μk is the coefficient of kinetic friction.

Kinetic friction acts opposite the direction of (relative) motion, therefore

The components along the X and Y directions are:


Case 2 – Static Friction

For this case, there is no relative motion between the bowling ball and the lane at point P. Therefore, VP = 0.

Using the same expression as before,

For pure rolling of the bowling ball the following condition must be satisfied:

where μs is the coefficient of static friction.


This completes the analysis of the physics of bowling.


Calculating the position angle of the moon (the rolling back and forth) - Astronomy

As the Earth rotates to the East, any object not attached to it will seem to drift to the West. This is why the stars, Sun, Moon and planets all rise in the East and set in the West. Their motion is, in a sense, an optical illusion, a mirror image of the motion which we actually have. Similarly, there are effects which act on objects near the surface of the Earth which are not so obvious as the rotation of the sky, but which can be directly verified by careful observation. These effects can be explained in terms of an 'imaginary' force called the Coriolis force (after Gustave Coriolis, a French engineer and mathematician who showed that such a force could be used to allow the use of the ordinary laws of motion in a rotating reference frame), and as a result the effects are referred to as Coriolis effects.
The nature and size of the Coriolis effects depend upon where you are. If you are near the Pole the axis of the Earth's rotation is nearly vertical relative to your Horizon, and as things spin around in the sky they move nearly horizontally. Similarly, the Coriolis effects are almost completely horizontal. But if you were near the Equator, where the Earth's axis of rotation is nearly horizontal, things spinning around the sky would move nearly vertically and the Coriolis effects are almost completely vertical. At in-between latitudes the stars rise and set at an angle to both the vertical and the horizontal, and there are both vertical and horizontal Coriolis effects. In general, as you move toward one of the Poles the horizontal Coriolis effects grow and the vertical ones shrink, whereas if you move toward the Equator the horizontal effects shrink and the vertical ones grow.

The vertical effect (predicted by Newton)
The vertical Coriolis effect was actually predicted by Newton when he discovered the laws which govern the motions of things throughout the Universe. As we move around the Earth we have some speed (in Long Beach, 870 mph to the East), and if no force were to act on us we would move in a straight line with that speed, without any change. But as the Earth rotates we have to move along a circle (our parallel of latitude) to stay on the surface of the Earth. Since this is not a straight line, we need a force which points toward the center of that circle (toward the axis of rotation of the Earth) in order to not float off into space and spiral away from the ground beneath us. As a result the force of gravity, which we usually think of as just holding us on the Earth, actually has to do two things: hold us on the Earth so that we don't drift away, and also squeeze us into the ground. It is only the latter effect which causes us to feel heavy, and if part of the force of gravity is 'used up' keeping us on the ground, then the remainder of the force, the part which produces our 'weight', will appear to be less than usual.
If we were at the North Pole we wouldn't going anywhere, so no force would be required to hold us on the Earth and our weight would be equal to our true gravitational weight. But if we move towards the Equator our motion around our parallel of latitude would be faster and faster, so we would need more and more force just to keep us on the ground (up to 1/3 of one percent of our weight at the Equator), and we would therefore seem to weigh less and less, even though our true weight would always be just about the same. This is the phenomenon experienced by astronauts in orbit, which makes them feel weightless. Their speed is so great that all of their weight is used up just keeping them in orbit, so there is nothing left to squeeze them downwards. You can experience this feeling yourself by jumping off a building. While you are falling all your weight is used in making you fall, so you feel weightless until you hit the ground. Of course you can tell that you are falling by looking at the ground, so you know that you aren't really weightless, but for an astronaut in orbit the speed around the Earth is so great that the round shape of the Earth makes the ground "fall" away from him just as fast as he is falling, so it doesn't actually look like he is falling.
Since at the Poles you wouldn't be going anywhere, and the force required to hold you on the ground is zero and the apparent pull of gravity is the actual pull, things would have a weight and would fall in exactly the way that they really do. But near the Equator you are going around at over 1000 mph, about 1/3% of your gravitational weight is used up just holding you on the ground, and you would seem to weigh and to fall only 99 2/3% as much and as fast as at the Poles. Careful laboratory measurements can easily show that this is correct.
To see how this works, suppose that you lie down and let someone put a 300 pound lead weight on top of you. Regardless of where you are, the weight would weigh 300 pounds. At the Poles, where none of the weight is 'used up' keeping it on the ground, you would feel 300 pounds pressing down on you. At the Equator, where 1 pound would be 'used up' keeping the weight going around the Earth instead of off into space, you would only feel 299 pounds pressing down on you. The actual weight is the same, but you would feel a difference of 1 pound because of the vertical Coriolis effect. This difference produces, somewhat surprisingly, a remarkable result. At the Equator the Earth bulges out by 1/3% of its radius, or 12 miles, compared to the Poles. The rocks beneath the Earth's surface feel the full weight of the rocks above them if they are below the Poles, but only 99 2/3% of the full true weight if they are near the Equator, and they are more compressed below the Poles, so that they take up less room, and less compressed near the Equator, so that they take up more room. The effects of this compression depend upon the response of the rocks to differences in weight and cannot be easily predicted, but over 300 years ago Newton was able to show that insofar as the rocks inside the Earth are 'elastic' (springy), or behave like a fluid (able to move according to what forces act on them), the Earth would bulge out by 1/3% of its radius at the Equator. It took nearly 200 years after Newton for measurement techniques to become precise enough to show, to a reasonable degree of accuracy, that the actual shape of the Earth is close to the predicted shape, but that is indeed the case.

The horizontal effect: The Coriolis Effect
In addition to this effect due to the vertical part of the rotational effect, there is a horizontal effect, which is usually explained by describing the motion of a pendulum.
If you set a pendulum to swinging, the only forces normally acting on it are the force of gravity, which is downwards, and the force of the string supporting the pendulum bob, which is up and to the side. These two forces define a vertical plane, and if no other forces act on the pendulum bob the pendulum should swing back and forth in this plane without veering to either side of the plane.
If you set a pendulum up like this at the North Pole, you can see that it does always swing in the same way by comparing the direction of swing to the direction of the stars. It will continue to swing back and forth between the same (stellar) directions, without any change, as long as there is no twisting due to the string or its support. But because the Earth is rotating, the stars seem to gradually turn towards the West during the day, and so the pendulum, keeping the same position relative to the stars, seems to also swing to the West. In reality neither the stars nor the pendulum are swinging to the West. The Earth is simply turning to the East underneath them. But if we "forget" this, then the motion of the stars and the pendulum seem perfectly real. Pendulums set up to demonstrate the Earth's rotation in this way are called Foucault pendulums, after the first man to do this.

At the Poles the pendulum seems to turn to the West once each time that the Earth turns under it, or every 23 hours 56 minutes. At the North Pole, where East is to your left and West is to your right, the pendulum turns around to the right. At the South Pole, where East is to your right and West is to your left, the pendulum turns around to the left. At other latitudes the speed at which the pendulum appears to turn is different because the top of the pendulum is being dragged around the Earth with the building it is attached to, so the situation is not as simple as at the Poles, but the direction of turning is still the same: to the right in the Northern Hemisphere, to the left in the Southern Hemisphere. Close to the Poles the rate of turning is close to once a day, but the rate decreases as you approach the Equator, until it disappears completely at that location.

In the case of the vertical Coriolis effect, as discussed above, there were two types of results:
(1) Laboratory Experiments: Things seem to fall slower and weigh less the nearer the Equator the laboratory is.
(2) Global Results: The Earth bulges at the Equator by 1/3% (12 miles radius).

Similarly, there are two types of horizontal results:
(1) Laboratory Experiments: Things thrown horizontally veer off to the right in the Northern Hemisphere, and off to the left in the Southern Hemisphere, faster near the Poles and slower near the Equator, so that Foucault pendulums gradually rotate to the right in the Northern Hemisphere, and to the left in the Southern Hemisphere, faster near the Poles and slower near the Equator.
(2) Global Results: The circulation of the atmosphere and oceans and local weather patterns have consistent effects due to the rotation of the Earth. In the Northern Hemisphere winds circle around low-pressure zones counter-clockwise, and around high-pressure zones clockwise. In the Southern Hemisphere, winds circle around in the opposite direction.

On a non-rotating Earth, if there were a region with lower than normal air pressure, winds would blow so as to carry air into that region, filling it with more and more air until pressures were normalized. But because of the Coriolis effect, winds blowing into such a region tend to circle around it, allowing the low pressure to persist for some time. In fact if the winds reach high enough speeds, a separate effect, called the Bernoulli effect, can actually intensify the low pressure, causing the winds to try to push in even more. But because they are rotating around the low pressure zone, if the winds push in closer they will have to speed up, just like an ice skater pulling in their arms spins faster. In this way the Coriolis effect transforms the initially random energy of the winds into an organized and in some cases remarkably fast motion (e.g., hurricanes).
There are similar effects in the general circulation of a planet's atmosphere. Because the Sun provides more heat near a planet's Equator than near its Poles, warm air tends to rise at the Equator, cold air tends to sink at the Poles, and a general circulation tends to arise where warm air carries heat towards the Poles and cold air carries a lack of heat towards the Equator. If a planet were small enough there could be a single circulation pattern like this, but even on the Earth there is too much space to cover, and three circulation patterns emerge in each hemisphere. Near the Equator and the Poles surface winds blow towards the Equator and upper atmosphere winds blow towards the Poles in Hadley cells, while at in-between latitudes the winds blow in the opposite directions, in Ferrell cells.

Near the Equator and the Poles, air moves toward the Equator at the surface and toward the Poles at altitude. At mid-latitudes air moves toward the Poles at the surface, and toward the Equator at altitude.

If the Earth were not rotating, these wind circulations would be essentially North and South, but because of the Coriolis effect winds veer somewhat to the right of their original direction in the Northern Hemisphere, and to left in the Southern Hemisphere, producing "normal" wind circulation patterns such as these:

Summary
Someone at the surface of the Earth sees phenomena called Coriolis effects as a result of the Earth's rotation.
The vertical effect is an apparent reduction in the force of gravity as we move from the Poles to the Equator, and a resulting change in the rate of fall of objects but actually, the force of gravity is approximately the same at every latitude, and if anything is greater at the Equator than at the Poles, because of the excess material wrapped around the bulge of the Equator.
The horizontal effect is an apparent veering to the right or left of a straight-line path by an object moving horizontally as viewed from space, the object is actually moving in as straight-line a path as possible, given the curvature of the Earth's surface.
In other words, Coriolis effects can be viewed as an illusion caused by the rotation of the Earth, rather than real phenomena. They appear perfectly real to an observer moving with the Earth's rotation, but illusory to an observer viewing the motion from a reference frame fixed in space.

Further Topics
There is a whole class of phenomena, such as Coriolis effects, which could be explained by forces which produce the observed effects since the effects are illusory, the forces invoked to explain them are also illusory. We refer to such forces as fictitious forces in the case of the Coriolis effect this fictitious force is called the Coriolis force. To an observer in space not taking part in the rotation of the Earth, the Coriolis force and its supposed effects do not exist but to an observer on the ground the effects seem perfectly real, and may be explained by this apparently real force.
(more coming in the next iteration of this page -- an explanation of inertial and accelerated frames of reference fictitious forces, including the Coriolis force and the force of gravity and more diagrams, to better explain the topics above)


The History of the Sextant

Talk given at the amphitheatre of the Physics Museum under the auspices of the Pro-Rector for Culture and the Committee for the Science Museum of The University of Coimbra, the 3 October 2000.

So what do navigators need to find their position on the earth's surface by observing the stars?

  1. They need an Almanac prepared by the astronomers to forecast precisely where the heavenly bodies, the sun, moon planets and selected navigational stars, are going to be, hour by hour, years into the future, relative to the observatory that prepared the almanac, Greenwich, England in modern times.
  2. They need a chronometer or some other means of telling the time back at the observatory that was the reference point for the data in the almanac,
  3. It is the cartographer's job to provide accurate charts so that navigators can establish their position in latitude and longitude or in reference to landmasses or the hazards of rocks and shoals.
  4. The navigators need a quick and easy mathematical method for reducing the data from their celestial observations to a position on the chart
  5. Finally, navigators need an angle-measuring instrument, a sextant, to measure the angle of the celestial body above a horizontal line of reference.

How do navigators use the stars, including our sun, the moon, and planets to find their way? Well, for at least two millennia, navigators have known how to determine their latitude — their position north or south of the equator. At the North Pole, which is 90 degrees latitude, Polaris (the North Star) is directly overhead at an altitude of 90 degrees. At the equator, which is zero degrees latitude, Polaris is on the horizon with zero degrees altitude. Between the equator and the North Pole, the angle of Polaris above the horizon is a direct measure of terrestrial latitude. If we were to go outside tonight and look in the northern sky, we would find Polaris at about 40 degrees 13 minutes altitude - the latitude of Coimbra.

In ancient times, the navigator who was planning to sail out of sight of land would simply measure the altitude of Polaris as he left homeport, in today’s terms measuring the latitude of home port. To return after a long voyage, he needed only to sail north or south, as appropriate, to bring Polaris to the altitude of home port, then turn left or right as as appropriate and "sail down the latitude," keeping Polaris at a constant angle.

The Arabs knew all about this technique. In early days, they used one or two fingers width, a thumb and little finger on an outstretched arm or an arrow held at arms length to sight the horizon at the lower end and Polaris at the upper.

Kamal

In later years, they used a simple device called a kamal to make the observation. The kamal shown here actually is a modern piece that I made, but it’s very much like the ones used a thousand years ago, and probably much earlier. Notice the knots in the cord attached to the carved mahogany transom. Before leaving homeport, the navigator would tie a knot in the cord so that, by holding it in his teeth, he could sight Polaris along the top of the transom and the horizon along the bottom.

To return to homeport, he would sail north or south as needed to bring Polaris to the altitude he’d observed when he left home, then sail down the latitude. Over time, Arab navigators started tying knots in the string at intervals of one issabah. The word issabah is Arabic for finger, and it denotes one degree 36 minutes, which was considered to be the width of a finger. They even developed a journal of different ports that recorded which knot on the kamal corresponded to the altitude of Polaris for each port they frequently visited.

Throughout antiquity, the Greeks and Arabs steadily advanced the science of astronomy and the art of astrology. About a thousand years ago, in the 10 th century, Arabs introduced Europe to two important astronomical instruments—the quadrant and the astrolabe.

Astronomers Astrolabe. Arabic astronomer's astrolabe made by Hajji Ali of Kerbala around 1790. It’s about 3 and one-half inches in diameter. It was used to find the time of rising and setting of the sun and the altitude of the sun and selected stars. Importantly, it was used to find the direction of Mecca for the devout Moslem's morning and evening prayers.

In the word "astrolabe" - "astro means ‘star’ and "labe" roughly translates as ‘to take’ or 'to find.'

The astronomer's beautiful, intricate and expensive astrolabe was the grandfather of the much simpler, easy to use mariner's quadrant and astrolabe. The mariner’s quadrant—a quarter of a circle made of wood or brass--came into widespread use for navigation around 1450, though its use can be traced back at least to the 1200s.

Mariner’s brass quadrant. The scale spans 90 degrees and is divided into whole degrees. A plumb bob establishes a vertical line of reference. The quadrant shown here is a replica of the type Columbus might have used on his voyages to the New World. This one is marked off at the latitudes of Lisbon, Cabo Verde and Serra Leoa, down near the Equator where Columbus is known to have visited.

The quadrant was a popular instrument with Portuguese explorers. Columbus would have marked the observed altitude of Polaris on his quadrant at selected ports of call just as the Arab seaman would tie a knot in the string of his kamal.

Alternatively, the navigator could record the altura, or altitude, of Polaris quantitatively in degrees at Lisbon and at other ports to which he might wish to return. It wasn’t long before lists of the alturas of many ports were published to guide the seafarer up and down the coasts of Europe and Africa.

During the 1400’s, Portuguese explorers were traveling south along the coast of Africa searching for a route to the orient. As a seafarer nears the equator heading south, Polaris disappears below the horizon. So, in southern seas, mariners had to have a different way of finding their latitude. Under orders from the Portuguese Prince Henry, The Navigator, by 1480, Portuguese astronomers had figured out how to determine latitude using the position of the sun as it moved north and south of the equator with the seasons, what we now call its "declination." In simple terms, the navigator could determine his altura, his latitude, by using his quadrant to take the altitude of the sun as it came to it’s greatest altitude at local apparent noon, and then making a simple correction for the position of the sun north or south of the equator according to the date.

The mariner’s quadrant was a major conceptual step forward in seagoing celestial navigation. Like the knots-in-a string method of the Arab kamal, the quadrant provided a quantitative measure, in degrees, of the altitude of Polaris or the sun, and related this number to a geographic position—the latitude--on the earth’s surface. But for all its utility, the quadrant had two major limitations: On a windy, rolling deck, it was hard to keep it exactly vertical in the plane of a heavenly body. And it was simply impossible to keep the wind from blowing the plumb bob off line.

A beautiful mariners’ astrolabe made in Lisbon by J. de Goes in 1608, now in the Museum of the History of Science, Florence, Italy

Mariner's astrolabes are now very rare and expensive - less than one hundred are known to survive and most of these are in poor condition having been recovered from ship wrecks.

The seagoing astrolabe was a simplified version of the much more sophisticated Middle Eastern astronomer’s astrolabe that we saw a moment ago. All the complex scales were eliminated, leaving only a simple circular scale marked off in degrees. A rotatable alidade carried sighting pinnules. Holding the instrument at eye level, the user could sight the star through the pinnules and read the star’s altitude from the point where the alidade crosses the scale.

Astrolabe in use.For a sun sight, the astrolabe was allowed to hang freely and the alidade was adjusted so that a ray of sunlight passed through the hole in the upper vane and fell precisely on the hole in the lower vane.

The astrolabe was popular for more than 200 years because it was reliable and easy to use under the frequently adverse conditions aboard ship.

A cross-staff. This one is a modern reproduction in the style popular with Dutch navigators in the eighteenth century.

The next step in the evolution of celestial navigation instruments was the cross-staff, a device resembling a Christian cross. Interestingly, its operating principle was the same as that of the kamal. The vertical piece, the transom or limb, slides along the staff so that the star can be sighted over the upper edge of the transom while the horizon is aligned with the bottom edge.

The Persian mathematician Avicenna wrote about a cross-staff in the eleventh century. The concept probably arrived in Europe when Levi ben Gerson, working in the Spanish school at Catalan in 1342, wrote about an instrument called a balestilla that he described as a being made from a "square stick" with a sliding transom.

A cross-staff in use. This drawing, from a Spanish book on navigation published in 1552, shows how the cross-staff was used to determine the altitude of Polaris. If you’ve ever heard the phrase "shooting the stars," it comes from the practice of holding a cross-staff up to the user’s eye with one hand, with the transom grasped in the other hand so that the person looks like an archer taking aim at the sun.

Early cross-staffs had only two pieces - the staff and one transom. Over time they became more elaborate. After 1650, most "modern" cross-staffs have four transoms of varying lengths. Each transom corresponds to the scale on one of the four sides of the staff. These scales mark off 90, 60, 30, and 10 degrees, respectively. In practice, the navigator used only one transom at a time.

The major problem with the cross-staff was that the observer had to look in two directions at once - along the bottom of the transom to the horizon and along the top of the transom to the sun or the star. A neat trick on a rolling deck!

Davis quadrant. Made by an English craftsman named Walter Henshaw in 1711. It’s made of rosewood with a diagonal scale on boxwood.

One of the most popular instruments of the seventeenth century was the Davis quadrant or back-staff. Captain John Davis conceived this instrument during his voyage to search for the Northwest Passage. It was described in his Seaman’s Secrets published in 1595. It was called a quadrant because it could measure up to 90 degrees, that is, a quarter of a circle. The observer determined the altitude of the sun by observing its shadow while simultaneously sighting the horizon. Relatively inexpensive and sturdy, with a proven track record, Davis quadrants remained popular for more than 150 years, even after much more sophisticated instruments using double-reflection optics were invented.

One of the major advantages of the Davis back-staff over the cross-staff was that the navigator had to look in only one direction to take the sight - through the slit in the horizon vane to the horizon while simultaneously aligning the shadow of the shadow vane with the slit in the horizon vane.

The major problem with back-sight instruments was that it was difficult if not impossible to sight the moon, the planets or the stars. Thus, toward the end of the 1600's and into the 1700's, the more inventive instrument makers were shifting their focus to optical systems based on mirrors and prisms that could be used to observe the nighttime celestial bodies.

The critical development was made independently and almost simultaneously by John Hadley in England and by Thomas Godfrey, a Philadelphia glazier, about 1731. The fundamental idea is to use of two mirrors to make a doubly reflecting instrument—the forerunner of the modern sextant.

Diagram of sextant

How does such an instrument work? How many of you have ever held a sextant in your hand? Hold the instrument vertically and point it toward the celestial body. Sight the horizon through an unsilvered portion of the horizon mirror. Adjust the index arm until the image of the sun or star, which has been reflected first by the index mirror and second by the silvered portion of the horizon mirror, appears to rest on the horizon. The altitude of the heavenly body can be read from the scale on the arc of the instrument’s frame.

Hadley's first doubly reflecting octants were made from solid sheets of brass. They were heavy and had a lot of wind resistance. Lighter wooden instruments that could be made larger, with scales easier to divide accurately and with less wind resistance quickly replaced them.

Early Hadley octant. This mahogany octant was made about 1760 by the famous London maker, George Adams.

Hadley' octant of 1731 was a major advancement over all previous designs and is still the basic design of the modern sextant. It was truly a "point and shoot" device. The observer looked at one place - the straight line of the horizon sighted through the horizon glass alongside the reflected image of the star. The sight was easy to align because the horizon and the star seemed to move together as the ship pitched and rolled.

We have seen how navigators could find their latitude for many centuries but ships, crews and valuable cargo were lost in shipwrecks because it was impossible to determine longitude. Throughout the seventeenth century and well into the eighteenth century, there was an ongoing press to develop techniques for determining longitude . The missing element was a way to measure time accurately. The clock makers were busy inventing ingenious mechanical devices while the astronomers were promoting a celestial method called "lunar distances". Think of the moon as the hand of a clock moving across a clock face represented by the other celestial bodies. Early in the 18 th century, the astronomers had developed a method for predicting the angular distance between the moon and the sun, the planets or selected stars. Using this technique, the navigator at sea could measure the angle between the moon and a celestial body, calculate the time at which the moon and the celestial body would be precisely at that angular distance and then compare the ship’s chronometer to the time back at the national observatory. Knowing the correct time, the navigator could now determine longitude. When the sun passes through the meridian here at Coimbra, the local solar time is 1200 noon and at that instant it is 1233 PM Greenwich Mean Time. Remembering that 15 degrees of longitude is equivalent to one hour of time gives us the longitude of 8 degrees, 15 minutes West of Greenwich. The lunar distance method of telling time was still being used into the early 1900’s when it was replaced by time by radio telegraph.

An octant measures angles up to 90 degrees and is ideally suited for observations of celestial bodies above the horizon. But greater angle range is needed for lunar distance observations. It was a simple matter to enlarge Hadley's octant, an eighth of a circle, to the sextant, a sixth of a circle, that could measure up to 120 degrees.

An early sextant by John Bird. The first sextant was produced by John Bird in 1759. This is a very early example of his work now in the Nederlands Scheepvaart Museum in Amsterdam. The frame is mahogany with an ivory scale. It is so large and heavy that it needed a support that fitted into a socket on the observers belt.
A brass sextant by Dollond. Here’s a fine brass sextant from the early nineteenth century by the master London instrument maker John Dollond.

In the first half of the eighteenth century there was a trend back to wooden frame octants and sextants to produce lighter instruments compared to those made of brass.

Ebony sextant. A very handsome example by H. Limbach of Hull of a sextant with an ebony frame. Ebony was used because of the dense wood's resistance to humidity. The scale and vernier were divided on ivory, or should we now say bone. The design was not successful because the wood tended to split over the long arc of a sextant.

Examples of sextant frame designs. A sample of variations in frame design. The challenge was to produce sextant frames that were light weight, low wind resistance and with a minimum change is dimensions with changes in temperature. As you can see, some of them are quite esthetically pleasing.

Ramsden pentant . To be correct, the instrument should be called a pentant, a fifth of a circle, rather than a sextant. This jewel is only 4 1/2 inches radius. The scale is divided on silver from minus 5 degrees to 155 degrees with each degree further divided in three to 20 arc minutes. As you can see, the scale is beveled at 45 degrees. Why set the scale at an angle to the frame - perhaps just to show that he could do it!

Probably the finest 18 th century instrument maker was the Englishman Jesse Ramsden. His specialty was accurate scale division. Here’s a small brass sextant that Ramsden made shortly before his death in 1800. Ramsden's major achievement was to invent a highly accurate "dividing engine"—the apparatus used to divide the scale into degrees and fractions of degrees. His design was considered so ingenious that the British Board of Longitude awarded Ramsden a prize of 615 pounds—in 18 th century terms, a small fortune. His "dividing engine" now resides in the Smithsonian Institution in Washington.

The development of more precise scale division was a milestone in instrument development. Certainly, it permitted more accurate observations but it also permitted smaller, lighter, more easily handled instruments. The sextant you see here is my all-time favorite.


3.6 Finding Velocity and Displacement from Acceleration

This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocity and Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity and acceleration using the derivative. By taking the derivative of the position function we found the velocity function, and likewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, we can work backward and calculate the velocity function from the acceleration function, and the position function from the velocity function.

Kinematic Equations from Integral Calculus

Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,

we can take the indefinite integral of both sides, finding

where C1 is a constant of integration. Since

Similarly, the time derivative of the position function is the velocity function,

Thus, we can use the same mathematical manipulations we just used and find

where C2 is a second constant of integration.

We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find

If the initial velocity is v(0) = v0, then

which is (Equation). Substituting this expression into (Figure) gives

Doing the integration, we find

so, C2 = x0. Substituting back into the equation for x(t), we finally have

Example

Motion of a Motorboat

A motorboat is traveling at a constant velocity of 5.0 m/s when it starts to decelerate to arrive at the dock. Its acceleration is

. (a) What is the velocity function of the motorboat? (b) At what time does the velocity reach zero? (c) What is the position function of the motorboat? (d) What is the displacement of the motorboat from the time it begins to decelerate to when the velocity is zero? (e) Graph the velocity and position functions.

Strategy

(a) To get the velocity function we must integrate and use initial conditions to find the constant of integration. (b) We set the velocity function equal to zero and solve for t. (c) Similarly, we must integrate to find the position function and use initial conditions to find the constant of integration. (d) Since the initial position is taken to be zero, we only have to evaluate the position function at

Solution

We take t = 0 to be the time when the boat starts to decelerate.

    From the functional form of the acceleration we can solve (Figure) to get v(t):

At t = 0 we have v(0) = 5.0 m/s = 0 + C1, so C1 = 5.0 m/s or

At t = 0, we set x(0) = 0 = x0, since we are only interested in the displacement from when the boat starts to decelerate. We have

Therefore, the equation for the position is

Figure 3.30 (a) Velocity of the motorboat as a function of time. The motorboat decreases its velocity to zero in 6.3 s. At times greater than this, velocity becomes negative—meaning, the boat is reversing direction. (b) Position of the motorboat as a function of time. At t = 6.3 s, the velocity is zero and the boat has stopped. At times greater than this, the velocity becomes negative—meaning, if the boat continues to move with the same acceleration, it reverses direction and heads back toward where it originated.

Significance

The acceleration function is linear in time so the integration involves simple polynomials. In (Figure), we see that if we extend the solution beyond the point when the velocity is zero, the velocity becomes negative and the boat reverses direction. This tells us that solutions can give us information outside our immediate interest and we should be careful when interpreting them.

Check Your Understanding

A particle starts from rest and has an acceleration function

. (a) What is the velocity function? (b) What is the position function? (c) When is the velocity zero?

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    The velocity function is the integral of the acceleration function plus a constant of integration. By (Figure),

Summary

  • Integral calculus gives us a more complete formulation of kinematics.
  • If acceleration a(t) is known, we can use integral calculus to derive expressions for velocity v(t) and position x(t).
  • If acceleration is constant, the integral equations reduce to (Figure) and (Figure) for motion with constant acceleration.

Key Equations

Conceptual Questions

When given the acceleration function, what additional information is needed to find the velocity function and position function?

Problems

The acceleration of a particle varies with time according to the equation

. Initially, the velocity and position are zero. (a) What is the velocity as a function of time? (b) What is the position as a function of time?

Between t = 0 and t = t0, a rocket moves straight upward with an acceleration given by

, where A and B are constants. (a) If x is in meters and t is in seconds, what are the units of A and B? (b) If the rocket starts from rest, how does the velocity vary between t = 0 and t = t0? (c) If its initial position is zero, what is the rocket’s position as a function of time during this same time interval?

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The velocity of a particle moving along the x-axis varies with time according to

, where A = 2 m/s, B = 0.25 m, and

. Determine the acceleration and position of the particle at t = 2.0 s and t = 5.0 s. Assume that

A particle at rest leaves the origin with its velocity increasing with time according to v(t) = 3.2t m/s. At 5.0 s, the particle’s velocity starts decreasing according to [16.0 – 1.5(t – 5.0)] m/s. This decrease continues until t = 11.0 s, after which the particle’s velocity remains constant at 7.0 m/s. (a) What is the acceleration of the particle as a function of time? (b) What is the position of the particle at t = 2.0 s, t = 7.0 s, and t = 12.0 s?

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Additional Problems

Professional baseball player Nolan Ryan could pitch a baseball at approximately 160.0 km/h. At that average velocity, how long did it take a ball thrown by Ryan to reach home plate, which is 18.4 m from the pitcher’s mound? Compare this with the average reaction time of a human to a visual stimulus, which is 0.25 s.

An airplane leaves Chicago and makes the 3000-km trip to Los Angeles in 5.0 h. A second plane leaves Chicago one-half hour later and arrives in Los Angeles at the same time. Compare the average velocities of the two planes. Ignore the curvature of Earth and the difference in altitude between the two cities.

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Take west to be the positive direction.

Unreasonable Results A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. If his average velocity is 24 km/h, how long did it take him to complete the trip? Is this a reasonable time?

An object has an acceleration of

. Determine the object’s velocities at

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A particle moves along the x-axis according to the equation

m. What are the velocity and acceleration at

A particle moving at constant acceleration has velocities of

s. What is the acceleration of the particle?

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A train is moving up a steep grade at constant velocity (see following figure) when its caboose breaks loose and starts rolling freely along the track. After 5.0 s, the caboose is 30 m behind the train. What is the acceleration of the caboose?

An electron is moving in a straight line with a velocity of

m/s. It enters a region 5.0 cm long where it undergoes an acceleration of

along the same straight line. (a) What is the electron’s velocity when it emerges from this region? b) How long does the electron take to cross the region?

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An ambulance driver is rushing a patient to the hospital. While traveling at 72 km/h, she notices the traffic light at the upcoming intersections has turned amber. To reach the intersection before the light turns red, she must travel 50 m in 2.0 s. (a) What minimum acceleration must the ambulance have to reach the intersection before the light turns red? (b) What is the speed of the ambulance when it reaches the intersection?

A motorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. Calculate (a) the acceleration of the motorcycle and (b) its velocity at the beginning and end of the 2-km trip.

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solve simultaneously to get

. Velocity at the end of the trip is

A cyclist travels from point A to point B in 10 min. During the first 2.0 min of her trip, she maintains a uniform acceleration of

. She then travels at constant velocity for the next 5.0 min. Next, she decelerates at a constant rate so that she comes to a rest at point B 3.0 min later. (a) Sketch the velocity-versus-time graph for the trip. (b) What is the acceleration during the last 3 min? (c) How far does the cyclist travel?

Two trains are moving at 30 m/s in opposite directions on the same track. The engineers see simultaneously that they are on a collision course and apply the brakes when they are 1000 m apart. Assuming both trains have the same acceleration, what must this acceleration be if the trains are to stop just short of colliding?

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A 10.0-m-long truck moving with a constant velocity of 97.0 km/h passes a 3.0-m-long car moving with a constant velocity of 80.0 km/h. How much time elapses between the moment the front of the truck is even with the back of the car and the moment the back of the truck is even with the front of the car?

A police car waits in hiding slightly off the highway. A speeding car is spotted by the police car doing 40 m/s. At the instant the speeding car passes the police car, the police car accelerates from rest at 4 m/s 2 to catch the speeding car. How long does it take the police car to catch the speeding car?

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Equation for the speeding car: This car has a constant velocity, which is the average velocity, and is not accelerating, so use the equation for displacement with

Equation for the police car: This car is accelerating, so use the equation for displacement with

, since the police car starts from rest:

Now we have an equation of motion for each car with a common parameter, which can be eliminated to find the solution. In this case, we solve for

. The speeding car has a constant velocity of 40 m/s, which is its average velocity. The acceleration of the police car is 4 m/s 2 . Evaluating t, the time for the police car to reach the speeding car, we have

Pablo is running in a half marathon at a velocity of 3 m/s. Another runner, Jacob, is 50 meters behind Pablo with the same velocity. Jacob begins to accelerate at 0.05 m/s 2 . (a) How long does it take Jacob to catch Pablo? (b) What is the distance covered by Jacob? (c) What is the final velocity of Jacob?

Unreasonable results A runner approaches the finish line and is 75 m away her average speed at this position is 8 m/s. She decelerates at this point at 0.5 m/s 2 . How long does it take her to cross the finish line from 75 m away? Is this reasonable?

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At this acceleration she comes to a full stop in

, but the distance covered is

, which is less than the distance she is away from the finish line, so she never finishes the race.
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An airplane accelerates at 5.0 m/s 2 for 30.0 s. During this time, it covers a distance of 10.0 km. What are the initial and final velocities of the airplane?

Compare the distance traveled of an object that undergoes a change in velocity that is twice its initial velocity with an object that changes its velocity by four times its initial velocity over the same time period. The accelerations of both objects are constant.

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An object is moving east with a constant velocity and is at position

. (a) With what acceleration must the object have for its total displacement to be zero at a later time t ? (b) What is the physical interpretation of the solution in the case for

A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 1.30 s to go past the window. What was the ball’s initial velocity?

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velocity at the bottom of the window.

A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms

(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

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Unreasonable results. A raindrop falls from a cloud 100 m above the ground. Neglect air resistance. What is the speed of the raindrop when it hits the ground? Is this a reasonable number?

Compare the time in the air of a basketball player who jumps 1.0 m vertically off the floor with that of a player who jumps 0.3 m vertically.

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Consider the players fall from rest at the height 1.0 m and 0.3 m.

Suppose that a person takes 0.5 s to react and move his hand to catch an object he has dropped. (a) How far does the object fall on Earth, where

(b) How far does the object fall on the Moon, where the acceleration due to gravity is 1/6 of that on Earth?

A hot-air balloon rises from ground level at a constant velocity of 3.0 m/s. One minute after liftoff, a sandbag is dropped accidentally from the balloon. Calculate (a) the time it takes for the sandbag to reach the ground and (b) the velocity of the sandbag when it hits the ground.

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taking the positive root
b.

(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

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A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms

(d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

An object is dropped from a roof of a building of height h. During the last second of its descent, it drops a distance h/3. Calculate the height of the building.

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, h = total height and time to drop to ground

in t – 1 seconds it drops 2/3h

t = 5.45 s and h = 145.5 m. Other root is less than 1 s. Check for t = 4.45 s

Challenge Problems

In a 100-m race, the winner is timed at 11.2 s. The second-place finisher’s time is 11.6 s. How far is the second-place finisher behind the winner when she crosses the finish line? Assume the velocity of each runner is constant throughout the race.

The position of a particle moving along the x-axis varies with time according to

m. Find (a) the velocity and acceleration of the particle as functions of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position.

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c. The slope of the position function is zero or the velocity is zero. There are two possible solutions: t = 0, which gives x = 0, or t = 10.0/12.0 = 0.83 s, which gives x = 1.16 m. The second answer is the correct choice d. 0.83 s (e) 1.16 m

A cyclist sprints at the end of a race to clinch a victory. She has an initial velocity of 11.5 m/s and accelerates at a rate of 0.500 m/s 2 for 7.00 s. (a) What is her final velocity? (b) The cyclist continues at this velocity to the finish line. If she is 300 m from the finish line when she starts to accelerate, how much time did she save? (c) The second-place winner was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. What was the difference in finish time in seconds between the winner and runner-up? How far back was the runner-up when the winner crossed the finish line?

In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 295.38 km/h. The one-way course was 8.00 km long. Acceleration rates are often described by the time it takes to reach 96.0 km/h from rest. If this time was 4.00 s and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

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