Question

A cord with negligible mass is wrapped around a pulley that is a uniform disk of mass

5.00 kg

and radius

0.300 m

and that can rotate without friction about its central axis. A

1.0 kg

bucket is attached at the free end of the cord hanging down from the pulley and then released at time

t = 0

The cord begins to unwrap from the pulley as the bucket descends. At

t = 5.00 s

, through how many rotations has the pulley turned (the bucket is still descending)?

Accepted Answer

It is given that for the pulley,

$m_{P} = 5 kg, r = 0.3 m .$

Consider the image representation of the given problem.

The weight of the bucket is the source of its acceleration and also the torque exerted on the pulley. Hence, we can write for the bucket

$m_{B} g - T = m_{B} a$ , where $T$ is the tension in the string.

This tension is responsible for the rotational motion of the pulley, it exerts torque on the pulley.

Hence, $T = \frac{τ}{r}$ .

Hence, we obtain $m_{B} g - \frac{τ}{r} = m_{B} a \Rightarrow m_{B} g = m_{B} a + \frac{τ}{r}$ .

This further gives $m_{B} g = m_{B} \frac{α}{r} + \frac{I α}{r}$ .

Re-arranging the given terms to obtain the expression of angular acceleration, we get,

$α = \frac{g}{r (1 + \frac{I}{m_{B} r^{2}})}$ .

And $m_{B} = 1 kg$ bucket is tied to the end of the massless cord wound on the pulley.

Now, the angular acceleration of the pulley is given by the expression $α = \frac{g}{r (1 + \frac{I}{m_{B} r^{2}})}$ .

Since the pulley is in the form of a disc, the same expression, substituting the formula of moment of inertia of the disc, can be written as

$α = \frac{g}{r (1 + \frac{\frac{1}{2} m_{P} r^{2}}{m_{B} r^{2}})}$ .

Substitute the given value to obtain,

$α = \frac{10}{0.3 (1 + \frac{\frac{1}{2} (5)}{(1)})} = 9.52 rad s^{- 2} = 3 π rad s^{- 2}$ .

Using the second Kinematical equation of rotational motion, we have

$θ = ω_{i} t + \frac{1}{2} α t^{2}$ .

This, on assuming that the pulley was initially at rest, gives

$θ = 0 + \frac{1}{2} (3 π) (5)^{2} = \frac{75}{2} π rad$ .

Therefore, the total number of rotations that the pulley performs in the first $5 s$ is

$n = \frac{(\frac{75 π}{2})}{2 π} = 18.75$ .

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