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)
✓
Answer
$1st \,case:$
We can write the equation of motion for circular motion as:
$\omega^{2}=\omega_{0}^{2}+2 a \theta_{n}$
Now, $\omega=\frac{\omega_{0}}{2}, \theta=36 \times 2 \pi$ (given)
So, $\frac{\omega_{0}^{2}}{4}=\omega_{o}^{2}+2 \alpha \times 36 \times 2 \pi$
So, $\alpha=\frac{-3 \omega_{0}^{2}}{4 \times 144 \pi}$
$2nd \,case:$
$0=\frac{\omega_{0}^{2}}{4}-2 \times \frac{3 \omega_{0}^{2}}{4 \times 144 \pi} \times \theta$
So, $\theta=24 \pi$
So, number of rotations made by the fan before coming to rest
$=\frac{24 \pi}{2 \pi}=12$
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