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Question: A weight of mass
$m=2.6 {\rm kg}$ is suspended via
a light inextensible cable which is wound around a pulley of mass
$M=6.4 {\rm kg}$
and radius
$b=0.4 {\rm m}$. Treating the pulley as a uniform disk, find the downward
acceleration of the weight and the tension in the cable. Assume that the cable
does not slip with respect to the pulley.

Answer: Let $v$ be the instantaneous downward velocity of the weight, $\omega$ the
instantaneous angular velocity of the pulley, and $T$ the tension in the cable.
Applying Newton’s second law to the vertical motion of the weight, we obtain

The angular equation of motion of the pulley is written

where $I$ is its moment of inertia, and $\tau$ is the torque acting
on the pulley. Now, the only force acting on the pulley (whose
line of action does not pass through the pulley’s axis of rotation) is the tension in the cable. The
torque associated with this force is the product of the tension, $T$, and the perpendicular
distance from the line of action of this force to the rotation axis, which is equal to the
radius, $b$, of the pulley. Hence,

If the cable does not slip with respect to the pulley, then its
downward velocity, $v$, must match the tangential velocity of the outer surface of the
pulley, $b \omega$. Thus,

It follows that

The above equations can be combined to give

Now, the moment of inertia of the pulley is
$I=(1/2) M b^2$. Hence,
the above expressions reduce to
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Next: Worked example 8.5: Hinged
Up: Rotational motion
Previous: Worked example 8.3: Moment
Richard Fitzpatrick
2006-02-02

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