Circular Motion

Rotational kinematics

We'll discuss the basics of rotational kinematics in this chapter; the kinematics equations for constant angular acceleration are discussed in Chapter 7. The basic quantities of rotational kinematics are angular position, angular displacement, angular velocity, and angular acceleration.

angular position … measured in degrees or radians

review of angle measure in degrees and radians; remember that the radian is a"unitless" unit

Remember that for a complete circle, the circumference is

and therefore the angle of a complete circle is

But we know that for a complete circle, = 360°. Thus,

Chapter 6: Circular Motion, Orbits, and GravityTuesday, September 17, 2013

10:00 PM

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Angular displacement and angular velocity

Connecting linear and angular kinematic quantities

Using these two conversion factors allows one to convert from degrees to radians, or vice versa. For example,

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Example: maximum speed for a car turning around a curve on a level road with friction

A car of mass 1200 kg moves around a curve on level ground that has aa radius of 20 m. Determine the maximum speed for which the car cansafely move around the curve. The coefficient of friction is 0.5.

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Example: banking angle for a highway curve

Determine the ideal banking angle for a highway curve that has a (horizontal) radius of 20 m. Suppose that the typical driving speed around the curve is about equal to the speed determined in the previous example.

Solution: There is no friction on the road; presumably it's very icy.

Thus, the maximum acceleration that friction between the tires and the road can produce is 4.9 m/s2.

Notice that the maximum safe speed through the curve is independent of the mass of the vehicle; thus, a speed limit sign can be used that is appropriate for all vehicles, whether they are light motorcycles or heavy transport trucks.

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Notice that the ideal banking angle is independent of the mass of the vehicle; this is nice. It means that one can design a banked highway that will be appropriate for all vehicles, no matter their mass, so it will be just as safe for light motorcycles and heavy transport trucks.

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Example: apparent weight for motion in a vertical circle

Consider a car going around a vertical "loop-the-loop" of radius 3 m. Determine the minimum speed the car needs to make it through the loop.

Alternatively: Consider a bucket of water spinning in a vertical circle and determine the minimum speed (or angular speed) so that the water does not fall out of the bucket.

Solution: Draw a free-body diagram for the car when it is at the top of the loop:

This means that if the car is to complete the circle, the force must be provided by the normal force from the loop and gravity. As the speed increases, the normal force has to increase to provide the necessary force. On the other hand, if the speed of the car decreases, then the normal force will also decrease, until at a critical speed, the weight of the car will be sufficient to provide the centripetal force. If the speed were to decrease below this critical minimum value, the car will leave the loop and crash down.

Thus, the minimum speed for the car to make it through the loop corresponds to n = 0. Setting n = 0 and solving for the speed of the car, we obtain:

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centrifuges•

Read about centrifuges in the text book; they provide a nice practical example of circular motion. (Also, you'll think about physics the next time you use a lettuce spinner, which is a sort of centrifuge.)

Newton's law of gravity

Example: gravitational force between Earth and Moon

This may not seem like a very high speed, but remember that the loop is not very big. I once saw a "cirque" stunt where motorcycles were flying around the inside of a spherical metal structure, and the radius might have been this big and the speeds seemed about this fast or a bit faster. For a much bigger loop, a larger speed is required.

Now solve the problem of the water in the bucket yourself. How fast do you have to swing a bucket around so that the water doesn't fall out?

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Example: gravitational force between Earth and a small object of mass m at the Earth's surface

Newton's law of gravity is an inverse-square force law, and has the same structure as Coulomb's law for the force between two charged particles at rest. The diagram above is intended to illustrate that the force decreases by a factor of 4 when the distance between the objects doubles.

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orbital motion of a satellite around Earth … direction of gravitational forces at various points of the orbit … gravitational acceleration is approximately constant near surface, but the direction is clearly not constant over larger scales, nor is the magnitude constant over larger scales

This provides insight into our assumption earlier in the course that the acceleration due to gravity g is constant; we can see by the previous equation that this is not exactly true. Close to the Earth's surface it is approximately true, but as you move away from the Earth's surface the value of the acceleration due to gravity decreases.

The equation above also gives us a way to "weigh" the Earth. The acceleration due to gravity can be measured in a laboratory (in fact you did so in the pendulum experiment in this course), and so can the gravitational constant G(look up the famous Cavendish experiment for details). The radius of the Earth can be determined using an ingenious geometrical method first devised by Eratosthenes (you can also look this up); then the previous equation can be solved for the mass of the Earth.

The same formula can be used to determine the acceleration due to gravity on other planets, moons, asteroids, etc. Just replace the mass and radius of Earth by the mass and radius of the other planet. Also note that some books call the acceleration due to gravity at the surface (i.e., "g") by the term "surface gravity."

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Weightlessness in space

satellites in orbit are in free fall … hence occupants are weightless … check the textbook for details

Kepler's third law of planetary motion

Using Newton's law of gravity and Newton's second law of motion, we can derive Kepler's third law of planetary motion.

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Example: Use Kepler's third law of planetary motion to determine the distance between the Earth and Sun, given that the mass of the Sun is about 2 × 1030 kg.

Solution: Make sure to convert the period of the Earth into seconds:

If the orbit of the planet is elliptical instead of circular, a more complex analysis shows that Kepler's third law is still valid provided that we use the "semi-major axis" of the orbit in place of r. The semi-major axis of the elliptical orbit is the distance from the centre of the ellipse to the most distant point on the orbit.

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Dark matter: One of today's unsolved puzzles about the universe

As we discussed in class, if you are deep below the surface of the Earth, let's say in a very deep mine shaft, your weight is less than at the surface of the Earth. Only the mass "interior" to you (i.e., at radii smaller than yours) is effective in exerting a force on you; the force exerted on you by the mass of the Earth that is at larger radii cancels. This means that if you were anywhere inside a hollow spherical shell of mass, provided the shell has constant density, the gravitational force on you is zero. (If you wish to learn more about this, look up "Gauss's law" for gravity (there is a version of Gauss's law for electrostatic forces as well); to understand the mathematical argument, you'll need to have some integral calculus under your belt.)

The same principle can be applied to the motions of stars in our galaxy. If you analyze the motion of stars at various positions in our galaxy, you can deduce the amount of mass in the galaxy that lies closer to the galactic centre than the given star (using Newton's law of gravity and Newton's laws of motion). Repeating this kind of analysis for many stars gives us a good idea for the distribution of mass in the galaxy.

And this leads to a puzzle: The amount of mass that we detect by usual means (regular light telescopes, radio telescopes, etc.) is nowhere near enough to account for the mass that we know must be there by analyzing motions in the galaxy. That is, the "visual matter" does not account for all the matter that must be present; there must be some "dark matter."

What on Earth can this dark matter be? Nobody knows. It is highly unlikely that it could be simply ordinary matter that can't be detected (such as "cold" dust particles or gas, abandoned TV sets, etc.), so scientists have turned to more speculative possibilities. Maybe dark matter is some exotic new form of matter. If so, such forms of matter have not been detected in laboratories, which leaves us no closer to resolving the puzzle.

This is an example of the type of unresolved puzzle that is found at the frontier of every branch of science. There are always unsolved puzzles, which means there is always room for new ideas, and

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Geostationary satellite orbits

It's convenient to have communications satellites that orbit Earth above its equator with a period equal to Earth's rotational period; in this way, they "hover" over the same geographical point on Earth. Using Kepler's third law we can calculate the radius of the orbit of such "geostationary" satellites.

This is the distance from the centre of the Earth, so the distance of such a satellite from the surface of the Earth is 6400 km less, which is 35,850 km above the Earth's surface.

The International Space Station orbits Earth at an altitude of about 400 km, which is considered "low Earth orbit;" geosynchronous satellites are in "high Earth orbit."

Additional exercises:

Page 196, Exercise 7: A turntable rotates counterclockwise at 78 rpm. A

puzzles, which means there is always room for new ideas, and creative researchers have plenty of opportunities for making interesting new discoveries. Maybe one of you will devote the time and work necessary to reach one of the research frontiers; it will take a lot of time and work to reach the frontier, but for the right kind of person (i.e., one who is persistent and willing to put up with a certain amount of failure and frustration) the journey will be a lot of fun and very satisfying.

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speck of dust on the turntable is at = 0.45 rad at t = 0 s. Determine the

angle of the speck at t = 8.0 s. (The result should be between 0 and 2.)

Solution:

Page 197, Exercise 27: A satellite orbiting the Moon very near the surface has a period of 110 min. Use this information, together with the radius of the Moon (which is 1.74 × 106 m), and the mass of the Moon (which is 7.36 × 1022 kg), to calculate the free-fall acceleration on the Moon's surface.

Solution:

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Page 197, Exercise 45: A 500 g ball swings in a vertical circle at the end of a 1.5-m-long string. When the ball is at the bottom of the circle, the tension in the string is 15 N. Determine the speed of the ball at this point.

Solution:

This means that the surface gravity (which is another word for the free-fall acceleration at the surface) on the Moon is about 1/6 as much as the surface gravity on the Earth. How would this change life for you if you lived on the Moon for a while?

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