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 4.10 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 = 45.6 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 = 22.0 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.

This answer has not been graded yet.

(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?

Choose:

1. higher

2. lower    

3. the same

4. The ball never makes it to the top of the loop.

(e) Explain your answer.

Explanation / Answer

Since, the ball is rolling without slipping, there is no friction between the ball and track. Hence, we can easily apply energy conservation to solve this problem

(a) KEi + PEi + REi = KEf + PEf + REf

=> (1/2)*m*(4.1)2 + 0 + (1/2)*[(2/3)*m*(R)2]*[(4.1/R)2]= (1/2)*m*(v)2 + m*g*(2*0.456) + (1/2)*[(2/3)*m*(R)2]*[(v/R)2]

=> 8.4 + 5.6 = (1/2)(v)2 + 8.95 + (1/3)(v)2

=> v = 2.46 m/s

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(b) Centripetal acceleration (ac) = (v)2/r = 13.27 m/s2

Since, ac is greater than g, the ball must be in contact with the track pushing down on the ball.

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(c) KEi + PEi + REi = KEf + PEf + REf

=> (1/2)*m*(4.1)2 + 0 + (1/2)*[(2/3)*m*(R)2]*[(4.1/R)2]= (1/2)*m*(v)2 - m*g*(0.22) + (1/2)*[(2/3)*m*(R)2]*[(v/R)2]

=> 8.4 + 5.6 = (1/2)(v)2 - 2.16 + (1/3)(v)2

=> v = 19.4 m/s

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(d) As the ball is not rolling anymore, we need not consider rotational energy.

Moreover, since friction is negligible, we can ignore the work done by it

KEi + PEi= KEf + PEf

=> (1/2)*m*(4.1)2 + 0 = (1/2)*m*(v)2 - m*g*(2*0.456)

=> 8.4 = (1/2)(v)2 + 8.95

=> v2 = -1.1

The above relation is not possible. Therefore, the ball never makes it to the top of the loop

option 4 is the correct answer

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