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ag
com
22
ms
=−
μ
021 98
21
..
.
bg
c
h
where the minus sign indicates that the center of mass acceleration points left, opposite to
its velocity, so that the ball is decelerating.
(c) Measured about the center of mass, the torque exerted on the ball due to the frictional
force is given by
τ
mgR
. Using Table 102(f) for the rotational inertia, the angular
acceleration becomes (using Eq. 1045)
()
2
2
2
5 0.21 9.8 m/s
5
47 rad s
2
5
2
2 0.11 m
mgR
g
Im
R
R
τμ
α
−
−−
==
=
=
=
−
where the minus sign indicates that the angular acceleration is clockwise, the same
direction as
ω
(so its angular motion is “speeding up’’).
(d) The centerofmass of the sliding ball decelerates from
v
com,0
to
v
com
during time
t
according to Eq. 211:
vv
g
t
com
com,0
. During this time, the angular speed of the ball
increases (in magnitude) from zero to
according to Eq. 1012:
ωα
=
t
gt
R
v
R
5
2
com
where we have made use of our part (a) result in the last equality. We have two equations
involving
v
com
, so we eliminate that variable and find
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 Spring '08
 Any
 Physics

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