A flywheel starts from rest and rotates at a constant acceleration of \(2 \mathrm{rad} \mathrm{s}^{-2}\).…

A flywheel starts from rest and rotates at a constant acceleration of \(2 \mathrm{rad} \mathrm{s}^{-2}\). The number of revolutions that it makes in first \(10 \mathrm{~s}\) is
  1. 16
  2. 24
  3. 32
  4. 8

Solution

Initial angular velocity of flywheel, \(\omega_0=0\) Angular acceleration, \(\alpha=2 \mathrm{rad} / \mathrm{s}^2\) Angular displacement in \(t=10 \mathrm{~s}\) is given as \(\begin{aligned} \theta & =\omega_0 t+\frac{1}{2} \alpha t^2 \\ & =0 \times 10+\frac{1}{2} \times 2 \times 10^2 \\ & =100 \mathrm{rad} \end{aligned}\) \(\begin{aligned} \text {Number of revolution } & =\frac{\theta}{2 \pi}=\frac{100}{2 \pi}=\frac{100}{2 \times \frac{22}{7}} \\ & =\frac{700}{44}=15.9 \simeq 16 \end{aligned}\)

Asked in: AP EAMCET 2020 (17 Sep Shift 1)

Practice more Rotational Motion questions on Aicharya