Question

A wheel rotates with the constant acceleration of $2.0\mathrm{rad}/{\mathrm{s}}^2$ , if the wheel starts from rest, then the number of revolutions it makes in the first $10\mathrm{s}$ will be approximately equal to

• A. 32
• B. 24
• C. 16
• D. 8

Answer 1

The correct option is C

16

Step 1:Given parameters

Angular acceleration, $\alpha =2.0\mathrm{rad}/{\mathrm{s}}^2$.

Time period, $t=10\text{s}$.

Step 2: Formula used:

For circular angular motion, the formula for angular displacement $\theta$ and angular acceleration$\alpha$

$\theta =\omega t+\frac{1}{2}\alpha t^2$

Where, $\omega$ is the initial angular velocity.

Step 2: Calculate the angular displacement

If $\theta$ be the angular displacement, then

$$\begin{gathered} \theta =\omega t+\frac{1}{2}\alpha t^2 \\ \theta =0\times t+\frac{1}{2}\left(2\right){\left(10\right)}^2 \\ \theta =100\mathrm{radians} \end{gathered}$$

Step 3: Calculate the number of revolutions

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{revolutions}=\frac{\theta }{2\mathrm{\pi }} \\ =\frac{100}{2\mathrm{\pi }} \\ =15.92\simeq 16 \end{gathered}$$

Therefore, the number of revolutions is $16$.

Hence, option (C) is correct.