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 4.14 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 = 45.9 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 = 24.0 cm below the horizontal section.

(a) Find the ball's speed at the top of the loop.

? m/s

(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.

? m/s

(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?

higher

lower

the same

the ball never makes it to the top of the loop

(e) Explain your answer.

Explanation / Answer

a) For a non-slipping hollow sphere,
KE = ½mv² + ½I² = ½mv² + ½(2mr²/3)(v/r)² = ½mv² + (1/3)mv² = (5/6)mv²
Then initial E = KE = (5/6) * M * (4.14m/s)² = M * 14.283 m²/s²
At the top of the loop, PE = mgh = M * 9.8m/s² * 0.918m = M * 9.0 m²/s²
Then the remaining KE = M * 5.283 m²/s² = (5/6) * M * v²
v = 2.52 m/s

b) V(min) = sqrt(rg) = sqrt(0.459*9.81) = 2.12 m/s

c) Vf = sqrt[((5/6)v^2 + hg)/(5/6)] = 4.47 m/s

d) and e) If you give it the same initial velocity but it is only sliding not rolling. Then it will have less energy than when rolling. As a result it will has less velocity at the top.

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