Question

A disc of mass ${{m}_{0}}$ rotates freely about a fixed horizontal axis through its centre. A thin cotton pad is fixed to its rim, which can absorb water. The mass of water dripping onto the pad is $\mu$ per second. After what time will the angular velocity of the disc get reduced to half of its initial value?
A. $\dfrac{2{{m}_{0}}}{\mu }$
B. $\dfrac{3{{m}_{0}}}{\mu }$
C. $\dfrac{{{m}_{0}}}{\mu }$
D. $\dfrac{{{m}_{0}}}{2\mu }$

Answer 1

Whenever the concept of rotation of bodies is involved, mass of the body must be replaced with its moment of inertia, velocity of the body must be replaced with its angular velocity, acceleration of the body must be replaced with its angular acceleration, so on and so forth. Also, the angular momentum remains conserved of the system if no external force is acting on it.

Complete step-by-step solution:
The moment of inertia of disc, $I=\dfrac{m{{R}^{2}}}{2}$
Mass of water dripping, $m=\mu t$
Where, $t=$ time

When the water droplets fall on the rim, the angular velocity of the fallen water droplets becomes equal to the angular velocity of the rim.

Let initial angular velocity of the body $={{\omega }_{0}}$

Given that the final angular velocity should get equal to half of the initial angular velocity, therefore, the final angular velocity of body, $\omega =\dfrac{1}{2}{{\omega }_{0}}$

By conservation of angular momentum, we get
Initial angular momentum = Final angular momentum

$\Rightarrow {{I}_{i}}{{\omega }_{i}}={{I}_{f}}{{\omega }_{f}}$
$\Rightarrow \left( \dfrac{1}{2}{{m}_{0}}{{R}^{2}} \right){{\omega }_{0}}={{m}_{w}}{{R}^{2}}\omega +\left( \dfrac{1}{2}{{m}_{0}}{{R}^{2}} \right)\omega$
Where,
${{m}_{w}}=$ mass of water $=\mu t$
$\begin{aligned} & \Rightarrow \left( \dfrac{1}{2}{{m}_{0}}{{R}^{2}} \right){{\omega }_{0}}=\left( \mu t \right)\left( {{R}^{2}} \right)\left( \dfrac{{{\omega }_{0}}}{2} \right)+\left( \dfrac{1}{2}{{m}_{0}}{{R}^{2}} \right)\left( \dfrac{{{\omega }_{0}}}{2} \right) \\\ & \Rightarrow {{m}_{0}}=\mu t+\dfrac{{{m}_{0}}}{2} \\\ & \Rightarrow {{m}_{0}}-\dfrac{{{m}_{0}}}{2}=\mu t \\\ & \Rightarrow \mu t=\dfrac{{{m}_{0}}}{2} \\\ \end{aligned}$
$\Rightarrow t=\dfrac{{{m}_{0}}}{2\mu }$

After time, $t=\dfrac{{{m}_{0}}}{2\mu }$, the angular velocity of the disc will get reduced to half to its initial value.

Therefore, the correct option is (D) $\dfrac{{{m}_{0}}}{2\mu }$.

Note:
Here, we have taken the mass of the water fallen on the cotton pad as a function of time because it is constantly changing with time as more and more water continues to fall gradually. Also, if a body is going under rotational motion, the velocity ($v$) of the body becomes equal to its radius times angular velocity, that is, $v=r\omega$