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 4.13 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 = 48.1 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 = 18.0 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? higher lower the same The ball never makes it to the top of the loop. (e) Explain your answer.

Explanation / Answer

v = 4.13 m/s ; r = 48.1 cm ; h = 18 cm

a)from the conservation of energy:

mghi + 1/2 m vi^2 + 1/2 I wi^2 = mghf + 1/2 m vf^2 + 1/2 I wf^2

I = 2/3 m r^2 ; v = r w=> w = v/r ; wi = vi/r ; wf = vf/r

vf = sqrt [{2g(hi - hf) + (1 + 2/3m r^2/m r^2)vi^2}/(1 + 2/3 mr^2/mr^2)]

vf = sqrt [(5/6vi^2 - 2gr)5/6]

vf = sqrt [(0.833 x 4.13^2 - 2 x 9.8 x 0.481)/0.833] = 2.39 m/s

Hence, v = 2.39 m/s

b)centripital acceleration of the ball is:

ac = vf^2/r = 2.39^2/0.481 = 11.88 m/s^2

and g = 9.81 m/s^2

since the centripital acceleration is greater than acceleration due to gravity, it will not fall from the loop.

c)The speed will be given by

1/2 m vi^2 + 1/2 i w^2 = 1/2 m vf^2 + 1/2 i w^2 - m g h

vf = sqrt [(5/6vi^2 + gh)/5/6]

vf = sqrt [0.833 x 4.13^2 + 9.8 x 0.18/0.833] = 4.38 m/s

Hence, vf = 4.38 m/s

d)In that case all energy will be just the translational kinetic energy, so the value of vf will be higher.

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