Problem 10.84 A child rolls a basketball of mass 0.640 kg up a long ramp. The ba
ID: 1403948 • Letter: P
Question
Problem 10.84
A child rolls a basketball of mass 0.640 kg up a long ramp. The basketball can be considered a thin-walled hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.60 m/s . When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.10 m/s . Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping.
Part A
Find the maximum vertical height increase of the ball as it rolls up the ramp.
Take free fall acceleration to be g = 9.80 m/s2 .
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Problem 10.84
A child rolls a basketball of mass 0.640 kg up a long ramp. The basketball can be considered a thin-walled hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.60 m/s . When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.10 m/s . Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping.
Part A
Find the maximum vertical height increase of the ball as it rolls up the ramp.
Take free fall acceleration to be g = 9.80 m/s2 .
hmax = mSubmitMy AnswersGive Up
Explanation / Answer
Initial velocity of the ball with which it is thrown (u) = 8.6m/s
Final velocity with which it comes back be (v) = 4.10m/s
Mass of ball (m) = 0.640 Kg
Total initial kinetic energy(k) is the sum of its rotational and translational kinetic energy -
Therefore,
k = (1/2)mu^2 + (1/2)Iw^2 ( where I is the moment of inertia of the ball and w, its angular velocity)
k = (1/2)mu^2 + (1/2) (2/3)mr^2(u/r)^2 ( I =(2/3)mr^2 for hollow sphere and w = u/r)
k = (5/6)mu^2
k = (5/6)* 0.640 * 8.6^2 J
k = 39.44J
Total Final kinetic enrgy (k') with which it comes back is given by-
k' = (5/6)mv^2
k' = (5/6) * 0.640 * 4.10^2
k' = 8.96 J
Loss of energy = k- k' = 39.44J - 8.96J = 30.48 J
This part of the energy is lost against friction and is the total work done by the friction.
As work done by Friction during Upward movement = work done by Friction during Downward movement = Wf
2 * Wf = 30.48 J
Wf = 15.24J
Applying work energy theorem. -
'Change in kinetic energy = work done on the particle by all the forces
Here only two forces are acting :- frictional force and gravity.
Therefore ,
0 - 39.44 J = - 15.24J - mgh
h= 24.2 / 0.640 * 9.8
h= 3.86 m
Maximum vertical height increase of the ball as it rolls up the ramp h = 3.86m