Worked example 8.4: Weight and pulley

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

$$m \dot{v} = m g-T.$$

The angular equation of motion of the pulley is written

$$I \dot{\omega} = \tau,$$

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,

$$\tau = T b.$$

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,

$$v = b \omega.$$

It follows that

$$\dot{v} = b \dot{\omega}.$$

The above equations can be combined to give

$$\dot{v} = \frac{g}{1+I/m b^2},$$

$$T = \frac{m g}{1+ m b^2/I}.$$

Now, the moment of inertia of the pulley is $I=(1/2) M b^2$. Hence, the above expressions reduce to

$$\dot{v} = \frac{g}{1+M/2 m} = \frac{9.81}{1+6.4/2\times 2.6}=4.40 {\rm m/s^2},$$

$$T = \frac{m g}{1+ 2 m/M}= \frac{2.6\times 9.81}{1+2\times 2.6/6.4} =14.07 {\rm N}.$$