A tennis ball is a hollow sphere with a thin wall. It is set rolling without sli

Question

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 3.98 m/s on a horizontal section of a track as shown in the figure below. It rolls around the inside of a vertical circular loop of radius r = 49.3 cm. As the ball nears the bottom of the loop, the shape of the track deviates from a perfect circle so that the ball leaves the track at a point h = 25.0 cm below the horizontal section. (a) Find the ball's speed at the top of the loop. (b) Demonstrate that the ball will not fall from the track at the top of the loop (c) Find its speed as it leaves the track at the bottom. (d) Suppose that static friction between ball and track were negligible so that the ball slid instead of rolling. Would its speed then be higher, lower, or the same at the top of the loop?

Explanation / Answer

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.15 m/s on a horizontal section of a track, as shown below. It rolls around the inside of a vertical circular loop 90.0 cm in diameter and finally leaves the track at a point 21.0 cm below the horizontal section. (a) Find the speed of the ball at the top of the loop. m/s (b) Find its speed as it leaves the track. m/s Since all energy is conserved, and there is no friction: a. Es = Et Es = Energy at start of track Et = Energy at top of track Ks + Us = Kt + Ut K = kinetic energy = .5mv^2 U = potential energy = mgh h_s = height at start h_t = height at top v_s = velocity at start v_t = velocity at top .5mv^2 + mgh_s = .5mv^2 + mgh_t .5v^2 + gh_s = .5v^2 + gh_t .5v_s^2 + gh_s = .5v_t^2 + gh_t v_s^2 + 2gh_s = v_t^2 + 2gh_t v_s^2 + 2gh_s - 2gh_t = v_t^2 v_s^2 + 2g(dh) = v_t^2 dh = Change in height of ball from start to top = -.9m The change in height is written as the height at start - height at top, which is negative diameter. v_s^2 + 2g(-.9m) = v_t^2 v_s^2 + 2g(-.9m) = v_t^2 (4.15 m/s)^2 + 2 * (9.8m/s^2) * (-.9m) = v_t^2 (v_t)^2 = -.4175m^2/s^2 v_t = .646i m/s What an imaginative velocity would indicate is that the ball would not reach the top of the the track; at least by conservation of energy arguments. This is because the change in gravitational potential energy is 8.82 * m, while the change in kinetic energy, if the ball were to go from 4.15m/s to 0m/s, is only 8.61125 * m; thus, the ball would not be able to reach the top of the track, assuming the given variables are correct. b. Assuming somehow the ball did get through the top of the track and went to the finish: Es = Ee Es = Energy at start Ee = Energy at end Ks + Us = Ke + Ue .5mv_s^2 + mgh_s = .5mv_e^2 + mgh_e .5v_s^2 + gh_s = .5v_e^2 + gh_e v_s^2 + 2gh_s = v_e^2 + 2gh_e v_s^2 + 2g(h_s - h_e) = v_e^2 Since the change in height (h_s - h_e) is .21m: (4.15m/s)^2 + 2(9.8m/s^2)(.21m) = v_e^2 = 21.3385m^2/s^2 v_e = 4.62 m/s Thus, the ball leaves with 4.62m/s of speed; this value makes sense given that the ball should leave the track faster than when it enters, since the ball will lose gravitational energy, and thus will gain some kinetic energy.

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