Part A If the ball is released from rest at a height of 0.51 m above the bottom
ID: 2033319 • Letter: P
Question
Part A
If the ball is released from rest at a height of 0.51 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track? Assume the ball is a solid sphere of radius 2.2 cm and mass 0.14 kg.(Figure 1)
Express your answer using two significant figures.
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Part B
How high does the ball rise on the frictionless side?
Express your answer using two significant figures.
SubmitRequest Answer
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1 of 1
Part A
If the ball is released from rest at a height of 0.51 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track? Assume the ball is a solid sphere of radius 2.2 cm and mass 0.14 kg.(Figure 1)
Express your answer using two significant figures.
? = rad/sSubmitRequest Answer
Part B
How high does the ball rise on the frictionless side?
Express your answer using two significant figures.
h = mSubmitRequest Answer
Return to AssignmentProvide Feedback
Figure1 of 1
No-slip FrictionlessExplanation / Answer
here,
height , h = 0.51 m
radius , r = 2.2 cm = 0.022 m
mass , m = 0.14 kg
A)
let the final angular speed at the bottom be w
using conservation of energy
0.5 * I * w^2 + 0.5 * m * v^2= m * g * h
0.5 * (2/5 * m * r^2) * w^2 + 0.5 * m * (r * w)^2 = m * g * h
0.7 * 0.022^2 * w^2 = 9.81 * 0.51
solving for w
w = 121.5 rad/s
B)
v = r * w = 2.67 m/s
let the ball will rise to a height h'
using conservation of energy
0.5 * m * v^2= m * g * h'
0.5 * 2.67^2 = 9.81 * h'
h' = 0.36 m
the ball will rise to a height 0.36 m