When a ceiling fan is switched off its angular velocity reduces to 50% while it makes 36 rotations. How many more rotation will it make before coming to rest (Assume uniform angular retardation)?

$\text{ By using equation }{\omega }^2={\omega }_0^2-2\alpha \theta$

${\left(\frac{{\omega }_0}{2}\right)}^2={\omega }_0^2-2\alpha \left(2\pi n\right)\Rightarrow \alpha =\frac{3}{4}\frac{{\omega }_0^2}{4\pi \times 36},(n=36) .......\left(\mathrm{i}\right)$

Now let fan completes total n' revolution from the starting to come to rest

$0={\omega }_0^2-2\alpha \left(2\pi n^{\mathrm{\prime }}\right)\Rightarrow n^{\mathrm{\prime }}=\frac{{\omega }_0^2}{4\alpha \pi }$

Substituting the value of $\alpha$ from equation (i)

$n^{\mathrm{\prime }}=\frac{{\omega }_0^2}{4\pi }\frac{4\times 4\pi \times 36}{3{\omega }_0^2}=48\text{ revolutions }$

Number of further rotations = 48 - 36 = 12