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Physics 8
Spring 2012
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Quiz 8 - Solutions
Make sure your name is on your quiz, and please box your final answer. Because we
will be giving partial credit, be sure to attempt all the problems, even if you don’t finish them!
Starting from rest, a hollow ball rolls without slipping down a ramp inclined at angle θ
to the horizontal. Find an expression for its speed, v, after it’s gone a distance d along the
incline.
Hint: Apply conservation of energy. Furthermore, the moment of inertia of a hollow ball
of mass M and radius R is I = 32 M R2 .
————————————————————————————————————
Solution
Consider the ball starting at rest at a height yi . It’s
initial energy is entirely potential energy, so Ei =
M gyi . As the ball rolls down the incline, as seen in
the figure to the right, it picks up speed and loses
potential energy. Since the ball is rolling it has
both translational and rotational kinetic energy.
So, the final energy is Ef = KEtrans + KErot +
M gyf = 21 M v 2 + 12 Iω 2 + M gyf , where v is the
translational speed, I is the moment of inertia, and
ω is the rotational speed. Conservation of energy
says that
H
d
θ
1
1
M g (yi − yf ) = M v 2 + Iω 2 .
2
2
But, the change in height H = yi − yf . Furthermore, from the geometry, H = d sin θ. Thus,
we find
1
1
M gd sin θ = M v 2 + Iω 2 .
2
2
Now, the condition that the ball roll without slipping means that v = Rω, where R is the
radius of the ball. Thus, we have
1
1 v2
M
I
2
M gd sin θ = M v + I 2 =
1+
v2.
2
2 R
2
M R2
Now, we are told that the moment of inertia is I = 32 M R2 , and so
M
2
5M 2
M gd sin θ =
1+
v2 =
v .
2
3
6
Canceling off the mass from both sides and solving for the speed gives
r
6
v=
gd sin θ.
5
1