A ping-pong player returns her opponent\'s seve with backspin. When it hits the

Question

A ping-pong player returns her opponent's seve with backspin. When it hits the table, it is moving forward at 5 m/s and rotating backwards at 6 revolutions per second.

1. The force of friction between the ball and the table is .05N. The ping pong ball is a spherical shell mass is 2.5g and radius 2cm.   Set up Newton's laws for rotational and linear acceleration.

2. How long does it take for the forward velocity to reach zero?

3. What is the angular velcoity at that time?

4. How much longer until the ball rolls without slipping, and in what direction does it roll?

Explanation / Answer

Wi = 6 rev/s

v = 5 m/s

u = 0.05

m = 2.5*10^3 kg

r = 2 cm = 0.02 m

So, force of friction = umg

So, acceleration,a = umg/m = ug

So, angular accelearion = a/r = ug/0.02

So, using the equation of motion, v = u +at

v= 0 <--- for stopping v=0

So, 0 = 5-0.05*9.8*t

So, t = 10.2 s <----- answer(2)

again,

Wf = Wi + At

Wf = 6-(0.05*9.8/0.02)*10.2

So, Wf = 243.9 rev/s <------answer(3)

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