When a ceiling fan is switched off, its angular velocity falls to half while it makes 36 rotations. How many…
- 24
- 36
- 18
- 12
Solution
$\begin{aligned} & \omega_f^2-\omega_i^2=2 a \theta \\ & \Rightarrow\left(\frac{\omega_f}{2}\right)^2-\omega_i^2=2 a \times 36 \times 2 \pi \quad ... (1) \end{aligned}$ For, angular velocity falls to zero when the fan is switched off,
\(\rightarrow \mathrm{O}^2-\omega^2=2 \mathrm{a} \times \mathrm{N} \times 2 \pi\) ...(2)
we get, \(N=48\)
So, the fan will make \(48-36=12\) more rotations before it stops.
Asked in: JEE Mains - Rotational Motion - Test 3