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

Ch6 Page 1 Chapter 6: Circular Motion, Orbits, and Gravity Tuesday, September 17, 2013 10:00 PM Circular Motion rotational kinematics angular position measured in degrees or radians review of angle measure in degrees and radians; remember that the radian is a "unitless" unit

Ch6 Page 2 angular displacement and angular velocity connecting linear and angular kinematic quantities

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Ch6 Page 5 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 a a radius of 20 m. Determine the maximum speed for which the car can safely move around the curve. The coefficient of friction is 0.5.

Ch6 Page 6 Thus, the maximum acceleration that friction between the tires and the road can produce is 4.9 m/s 2. 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. 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.

Ch6 Page 7 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. 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

Ch6 Page 8 minimum speed (or angular speed) so that the water does not fall out of the bucket. 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: This may not seem like a very high speed, but remember that the loop is not very big. I

Ch6 Page 9 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? 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

Ch6 Page 10 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. Example: gravitational force between Earth and a small object of mass m at the Earth's surface

Ch6 Page 11 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." 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 Weightlessness in space satellites in orbit are in free fall hence occupants are weightless check the textbook for

Ch6 Page 12 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. 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 10 30 kg. Solution: Make sure to convert the period of the Earth into seconds:

Ch6 Page 13 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." Ch6 Page 14