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12.67. Model:
The hollow cylinder is a rigid rotating body.
Visualize:
We placed the origin of the coordinate system on the ground.
Solve:
(a)
Newton’s second law for the block is
GB
B
()
,
y
FT
m
a
−+
=
where
T
is the tension in the string,
B
Fm
g
=
is the gravitational force on the block, and
y
a
is the acceleration of block. The string tension
exerts a negative (cw) torque on the cylinder, so the rotational form of Newton’s second law for the hollow
cylinder is
2
y
y
aI
a
TR
I
I
T
R
R
τα
=−
=
=
⇒ =−
where we used the acceleration constraint
.
y
aR
α
=
With this expression for
T
, Newton’s second law for the
block becomes
B
BB
22
B
(/
)
yy
Ia
m g
mg
ma
a
R
mI
R
−
−−
=
⇒
=
+
The moment of inertia of a hollow cylinder is
2
C
,
Im
R
=
so the equation for
y
a
simplifies to
2
2
B
BC
(3.0 kg)(9.8 m/s )
5.88 m/s
3.0 kg
2.0 kg
y
a
mm
==
=
−
++
The speed of the block just before it hits the ground can now be found using kinematics:
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 Fall '09
 geller

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