Gravitation and Circular Motion Practice Problems for AP Physics B & C (page 2)
Review the following concepts if necessary:
- Gravitation and Circular Motion for AP Physics B & C
- Newton's Law of Gravitation for AP Physics B & C
- Gravitation and Circular Motion: Of Special Interest to Physics C Students
Questions 1 and 2
Two stars, each of mass M, form a binary system. The stars orbit about a point a distance R from the center of each star, as shown in the diagram above. The stars themselves each have radius r.
- What is the force each star exerts on the other?
- In terms of each star's tangential speed v, what is the centripetal acceleration of each star?
- What force provides the rock's centripetal acceleration?
- The vertical component of the string's tension
- The horizontal component of the string's tension
- The entire tension of the string
- The gravitational force on the rock
- The horizontal component of the gravitational force on the rock
- The Ewok whirls the rock and releases it from a point above his head and to his right. The rock initially goes straightforward. Which of the following describes the subsequent motion of the rock?
- It will continue in a straight line forward, while falling due to gravity.
- It will continue forward but curve to the right, while falling due to gravity.
- It will continue forward but curve to the left, while falling due to gravity.
- It will fall straight down to the ground.
- It will curve back toward the Ewok and hit him in the head.
- A Space Shuttle orbits Earth 300 km above the surface. Why can't the Shuttle orbit 10 km above Earth?
- The Space Shuttle cannot go fast enough to maintain such an orbit.
- Kepler's laws forbid an orbit so close to the surface of the Earth.
- Because r appears in the denominator of Newton's law of gravitation, the force of gravity is much larger closer to the Earth; this force is too strong to allow such an orbit.
- The closer orbit would likely crash into a large mountain such as Everest because of its elliptical nature.
- Much of the Shuttle's kinetic energy would be dissipated as heat in the atmosphere, degrading the orbit.
- Consider two points on a rotating turntable: Point A is very close to the center of rotation, while point B is on the outer rim of the turntable. Both points are shown above. A penny could be placed on the turntable at either point A or point B.
- In which case would the speed of the penny be greater, if it were placed at point A, or if it were placed at point B? Explain.
- At which point would the penny require the larger centripetal force to remain in place? Justify your answer.
- Point B is 0.25 m from the center of rotation. If the coefficient of friction between the penny and the turntable is μ = 0.30, calculate the maximum linear speed the penny can have there and still remain in circular motion.
Questions 3 and 4: In the movie Return of the Jedi, the Ewoks throw rocks using a circular motion device. A rock is attached to a string. An Ewok whirls the rock in a horizontal circle above his head, then lets go, sending the rock careening into the head of an unsuspecting stormtrooper.
- D—In Newton's law of gravitation,
- E—In the centripetal acceleration equation
- B—Consider the vertical forces acting on the rock. The rock has weight, so mg acts down. However, because the rock isn't falling down, something must counteract the weight. That something is the vertical component of the rope's tension. The rope must not be perfectly horizontal, then. Because the circle is horizontal, the centripetal force must be horizontal as well. The only horizontal force here is the horizontal component of the tension. (Gravity acts down, last we checked, and so cannot have a horizontal component.)
- A—Once the Ewok lets go, no forces (other than gravity) act on the rock. So, by Newton's first law, the rock continues in a straight line. Of course, the rock still must fall because of gravity. (The Ewok in the movie who got hit in the head forgot to let go of the string.)
- E—A circular orbit is allowed at any distance from a planet, as long as the satellite moves fast enough. At 300 km above the surface Earth's atmosphere is practically nonexistent. At 10 km, though, the atmospheric friction would quickly cause the shuttle to slow down.
- Both positions take the same time to make a full revolution. But point B must go farther in that same time, so the penny must have bigger speed at point B.
- The coin needs more centripetal force at point B. The centripetal force is equal to mv2/r. However, the speed itself depends on the radius of the motion, as shown in part (a). The speed of a point at radius r is the circumference divided by the time for one rotation T, v = 2πr /T. So the net force equation becomes, after algebraic simplification, Fnet = 4mπ2r/T2. Because both positions take equal times to make a rotation, the coin with the larger distance from the center needs more centripetal force.
- The force of friction provides the centripetal force here, and is equal to μ times the normal force. Because the only forces acting vertically are FN and mg, FN = mg. The net force is equal to mv2/r, and also to the friction force μmg. Setting these equal and solving for v,
Plug in the values given (r = 0.25 m, μ = 0.30) to get v = 0.87 m/s. If the speed is faster than this, then the centripetal force necessary to keep the penny rotating increases, and friction can no longer provide that force.
the distance used is the distance between the centers of the planets; here that distance is 2R. But the denominator is squared, so (2R)2 = 4R2 in the denominator here.
the distance used is the radius of the circular motion. Here, because the planets orbit around a point right in between them, this distance is simply R.
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