Question

A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes ${{\left( \dfrac{1}{4} \right)}^{th}}$ of the original in time $'t'$ and $'n'$ revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are (Angular retardation is uniform)
(A). $\dfrac{4n}{15}$
(B). $\dfrac{8n}{15}$
(C). $\dfrac{16n}{15}$
(D). $\dfrac{32n}{15}$

Answer 1

A fan spinning about its axis only has circular motion. When the switch is turned off, the fan starts decelerating with a constant deceleration. We can apply equations of motion. Substituting corresponding values in the required equation of motion, we can calculate the number of revolutions completed.

Formula used:
$\omega ={{\omega }_{0}}+\alpha t$
${{\omega }^{2}}=\omega _{0}^{2}+2\alpha \theta$

Complete step by step solution:
Given that the fan decelerates with a constant deceleration, if initial velocity is $\omega$, final velocity is $\dfrac{\omega }{4}$ in time $'t'$ and $'n'$ revolutions are completed.
Using the following equation of motion,
$\omega ={{\omega }_{0}}+\alpha t$ - (1)
Here, $\omega$ is the initial angular velocity
${{\omega }_{0}}$ is the final angular velocity
$\alpha$ is the angular acceleration
$t$ is the time taken
Substituting given values in the above equation, we get,
$\begin{aligned} & \dfrac{\omega }{4}=\omega -\alpha t \\\ & \Rightarrow \alpha t=\omega -\dfrac{\omega }{4} \\\ & \Rightarrow \alpha t=\dfrac{3\omega }{4} \\\ & \therefore \alpha =\dfrac{3\omega }{4t} \\\ \end{aligned}$
The constant deceleration of the fan is $\dfrac{3\omega }{4t}$.
When the fun stops completely, final velocity is zero. The time taken will be-
Substituting given values in eq (1), we get,
$\begin{aligned} & 0=\omega -\dfrac{3\omega }{4t}\times t' \\\ & \Rightarrow \dfrac{3\omega }{4t}\times t'=\omega \\\ & \therefore t'=\dfrac{4t}{3} \\\ \end{aligned}$
The time taken by the fan to completely stop is $\dfrac{4t}{3}$.
Using the following equation of motion,
${{\omega }^{2}}=\omega _{0}^{2}+2\alpha \theta$ - (2)
Here, $\theta$ is the angular displacement covered by the fan
Substituting given values for when the fan stops completely in the above equation, we get,
$\begin{aligned} & 0={{\omega }^{2}}-2\times \dfrac{3\omega }{4t}\theta \\\ & \Rightarrow 2\times \dfrac{3\omega }{4t}\theta ={{\omega }^{2}} \\\ & \Rightarrow \theta =\dfrac{4t\omega }{6} \\\ \end{aligned}$
Therefore, the total angular displacement covered by the fan until it comes to rest is $\dfrac{4t\omega }{6}$.
Now, substituting values when initial velocity is $\omega$, final velocity is $\dfrac{\omega }{4}$ in time $'t'$ and $'n'$ revolutions are completed in eq (2), we get,
$\begin{aligned} & {{\left( \dfrac{\omega }{4} \right)}^{2}}={{\omega }^{2}}-2\times \dfrac{3\omega }{4t}n \\\ & \Rightarrow 2\times \dfrac{3\omega }{4t}n={{\omega }^{2}}-\dfrac{{{\omega }^{2}}}{16} \\\ & \Rightarrow 2\times \dfrac{3\omega }{4t}n=\dfrac{15{{\omega }^{2}}}{16} \\\ & \Rightarrow \omega =\dfrac{8n}{5t} \\\ \end{aligned}$
Therefore, substituting the value of $\omega$ in the number of revolutions made by the fan until it completely stops is-
$\dfrac{4t\omega }{6}=\dfrac{4t}{6}\times \dfrac{8n}{5t}=\dfrac{16n}{15}$
Therefore, the total number of revolutions completed by the fan until it completely stops is $\dfrac{16n}{15}$. Hence, the correct option is (C).

Note: The equations of motion in circular motion are analogous to equations of motion in a straight line. The acceleration is negative when a body is coming to rest. The equations of motion can be applied only when the acceleration of the body is constant. The angle covered in one revolution is the total angle of a circle.