A tennis ball is a hollow sphere (1=2/3 MR^2) with a thin wall. It is set rollin

Question

A tennis ball is a hollow sphere (1=2/3 MR^2) with a thin wall. It is set rolling without slipping at 4 .08 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 = 47.8 cm List all the initial energies for the ball and all the final (top of loop) energies for the ball. Some of these may just be formulas at this point. Find the ball's speed at the top of the loop. Suppose that static friction between ball and track were negligible so that the ball slid (no kinetic friction) instead of rolling. Would its speed then be higher, lower, or the same at the top of the loop? Explain.

Explanation / Answer

Here,

moment of inertia of ball, Ib = (2/3)*M*R^2

radius of loop, r = 47.8 cm = 0.478 m
initial velocity, vi = 4.08 m/s

Part A:
Ei = initial Energy
Ei = Kei + EKE
Ei = 0.5*M*v^2 + 0.5*I*w^2
Ei = 0.5*mv^2 + 0.5*(2mr^2/3)(v/r)^2
Ei = 0.5*mv^2 + (1/3)mv^2
Ei = (5/6)mv^2

At the top of the loop,
PE = mg(2r)

Part B :

Ei = (5/6)m*4.08^2
Ei = 13.872 * m

at top, PE = mg(2r)
PE = m*9.8*2*0.478
PE = 9.368 * m

The ramaining Kinectic erngy will be :

KE = (13.872 * m - 9.368 * m) = (5/6)mv^2

Solving for velocity at top, v

v = sqrt( (13.872 - 9.368) (6/5) )

v = 2.32 m/s

Part :
When the ball simply skid so, RKE = 0, Therefore

Ei = 0.5 * m *v^2

at top,
Pe = mg*2r
PE = 9.368 * m

Therefore
0.5 * m *v^2 = 9.368 * m

v = 4.33 m/s

The velocity will be doubled or can say increased

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