Question

A constant power is supplied to a rotating disc. The relationship between the angular velocity ($\omega $) of the disc and the number of rotations n made by the disc is governed by:
A) $\omega \propto {{n}^{\dfrac{1}{3}}}$
B) $\omega \propto {{n}^{\dfrac{2}{3}}}$
C)$\omega \propto {{n}^{\dfrac{3}{2}}}$
D)$\omega \propto {{n}^{2}}$

Answer 1

Hint: In the above question it is given that the disc is supplied with constant power. Hence the disc will rotate with a constant rate of supply of energy such that the angular velocity will keep on increasing. Hence equating the energy of a rotating body in terms of power and the energy of rotation will enable us to determine the required relation between angular velocity and the number of rotations of the disc.

Formula used:
$E=P\times t$
$E=\dfrac{1}{2}I{{\omega }^{2}}$

Complete step-by-step answer:
Let us say the above disc rotates with angular velocity $\omega $ at time ‘t’. If the disc is supplied with constant power ‘P’, then the energy ‘E’ acquired by the disc is given by,
$E=P\times t.....(1)$
In the above question it is given that the disc rotates. Hence the power supplied to the disc will get converted to the energy of rotation. If ‘I’ is the moment of inertia of the rotating disc, then the energy of rotation is given by,
$E=\dfrac{1}{2}I{{\omega }^{2}}....(2)$
Equating equation 1 and 2 we get,
$P\times t=\dfrac{1}{2}I{{\omega }^{2}}$
Differentiating the above equation by time we get,
$\begin{align}
  & \dfrac{d\left( P\times t \right)}{dt}=\dfrac{d\left( \dfrac{1}{2}I{{\omega }^{2}} \right)}{dt} \\
 & \Rightarrow P\dfrac{dt}{dt}=\dfrac{1}{2}I\dfrac{d{{\omega }^{2}}}{dt}=\dfrac{2\omega }{2}I\dfrac{d\omega }{dt} \\
 & \therefore \dfrac{d\omega }{dt}=\dfrac{P}{I\omega } \\
\end{align}$
Now in order to express the above equation in terms of the number of turns of the disc, let us multiply and divide the above equation by small change in angular position $d\theta $. Hence the above equation reduces to,
$\begin{align}
  & \dfrac{d\omega }{dt}\dfrac{d\theta }{d\theta }=\dfrac{P}{I\omega } \\
 & \because \dfrac{d\theta }{dt}=\omega \\
 & \Rightarrow \omega \dfrac{d\omega }{d\theta }=\dfrac{P}{I\omega } \\
 & \therefore {{\omega }^{2}}d\omega =\dfrac{Pd\theta }{I} \\
\end{align}$
Let us say the angular velocity of the disc is $\omega $ when the angular position is equal to $2n\pi $ where n is the number of rotations of the disc. Hence integrating the above equation we ge,
$\begin{align}
  & \int\limits_{0}^{\omega }{{{\omega }^{2}}d\omega }=\int\limits_{\theta =0}^{\theta =2n\pi }{\dfrac{Pd\theta }{I}} \\
 & \Rightarrow \dfrac{{{\omega }^{3}}}{3}=2n\pi \dfrac{P}{I} \\
 & \Rightarrow {{\omega }^{3}}=6n\pi \dfrac{P}{I} \\
 & \Rightarrow \omega ={{n}^{\dfrac{1}{3}}}{{\left( 6\pi \dfrac{P}{I} \right)}^{\dfrac{1}{3}}} \\
 & \therefore \omega \propto {{n}^{\dfrac{1}{3}}} \\
\end{align}$
Therefore the correct answer of the above question is option a.

So, the correct answer is “Option A”.

Note: It is to be noted that the above disc is not under pure rolling. Hence the energy supplied gets converted to the kinetic energy of rotation of the disc. The angular displacement for a single rotation is $2\pi $ and hence for ‘n’ rotations it's $2n\pi $.