Chapter 11
Problem 57
†
Given
ml = 0.440 kg
rl = 0.035 m
ωl = 180 rpm
mu = 0.270 kg
ru = 0.023 m
ωu = 0 rpm
Solution
a) Find the final speed of the two disks when they come to a common speed.
Since we are working with angular momentum for this problem, it will not hurt to remain in rpm’s for
the angular velocity. The answer will, therefore, be in rpm’s.
The moments of inertia for the disks are given by the relationship I = 12 m · r2 . By conservation of
angular momentum the initial angular momentum should equal the final angular momentum.
Li = Lf
Iu ωu + Il ωl = (Iu + Il )ωf
Solving for the final angular momentum gives
1
m r2 ω + 12 ml rl2 ωl
Iu ωu + Il ωl
2 u u u
ωf =
= 1
Iu + Il
m r2 + 21 ml rl2
2 u u
ωf =
1
(0.27
2
kg)(0.023 m)2 (0) + 12 (0.44 kg)(0.035 m)2 (180 rpm)
1
(0.27 kg)(0.023 m)2 + 21 (0.44 kg)(0.035 m)2
2
ωf = 142 rpm
b) Find the fraction of initial kinetic energy lost.
∆K
Kf − Ki
Kf
loss =
=
=− 1−
Ki
Ki
Ki
!
1
2
(I
+
I
)ω
(Iu + Il )ωf2
u
l
f
2
loss = − 1 −
=− 1−
1
Il ωl2
I ω2
2 l l
!
(mu ru2 + ml rl2 )ωf2
( 12 mu ru2 + 21 ml rl2 )ωf2
loss = − 1 −
=− 1−
1
ml rl2 ωl2
m r2 ω 2
2 l l l
((0.27 kg)(0.023 m)2 + (0.44 kg)(0.035 m)2 )(143 rpm)2
loss = − 1 −
(0.44 kg)(0.035 m)2 (180 rpm)2
loss = −0.202
% loss = −20.2 %
†
Problem from Essential University Physics, Wolfson