Question

When a ceiling fan is switched off, its angular velocity falls to half, while it makes $36$ rotations. How many more rotations will it make, before coming to rest?

Assume uniform angular retardation throughout the motion.

• A. $36$
• B. $24$
• C. $18$
• D. $12$

Answer 1

The correct option is D $12$

Number of rotations before its angular velocity falls to half when the fan is switched off,

$N=36$

So, total angular displacement,

$\theta =36\times 2\pi =72\pi$

Now,

${\omega }_f^2-{\omega }_i^2=2\alpha \theta ...\left(1\right)$

$\Rightarrow {\left(\frac{\omega }{2}\right)}^2-{\omega }^2=2\alpha \times 72\pi$

$\Rightarrow \alpha =-\frac{{\omega }^2}{192\pi }$

Further, again using equation $\left(1\right)$, to find the total angular displacement before the fan stops.

$\Rightarrow 0^2-{\omega }^2=2\times (-\frac{{\omega }^2}{192\pi })\times \theta$

$\Rightarrow \theta =96\pi$

Therefore, number of rotations,

$N^{\prime }=\frac{96\pi }{2\pi }=48$

So, the fan will make $48-36=12$ more rotations before it stops.

Hence, option $\left(D\right)$ is the correct answer.

$\text{Alternate Solution :}$

For, angular velocity falls to half when the fan is switched off,

${\omega }_f^2-{\omega }_i^2=2\alpha \theta$

$\Rightarrow {\left(\frac{\omega }{2}\right)}^2-{\omega }^2=2\alpha \times 36\times 2\pi ...\left(1\right)$

For, angular velocity falls to zero when the fan is switched off,

$\Rightarrow 0^2-{\omega }^2=2\alpha \times N\times 2\pi ...\left(2\right)$

On dividing equation $\left(1\right)$ by $\left(2\right)$ and solving, we get,

$N=48$

So, the fan will make $48-36=12$ more rotations before it stops.