Question

A wheel of radius $2\text{m}$ is rotating at a constant angular acceleration of magnitude $10{\text{rad/s}}^2$. Its angular frequency, at $t=0$, is $\frac{60}{\pi }\text{rpm}$. What will be its angular displacement at $t=2.0\text{sec}$?

• A. $\text{10 radian}$
• B. $\text{12 radian}$
• C. $\text{18 radian}$
• D. $\text{24 radian}$

Answer 1

The correct option is D $\text{24 radian}$

Given,

$R=2\text{m};\alpha =10{\text{rad/s}}^2$

Angular frequency, $f_0=\frac{60}{\pi }\text{rpm}$

$\therefore$ Angular speed

${\omega }_0=2\pi f_0=2\pi \times \frac{60}{\pi }=120\text{rad/minute}$

$\Rightarrow {\omega }_0=\frac{120}{60}\text{rad/s}=2\text{rad/s}$

At $t=2.0\text{sec}$, let the angular displacement is $\theta$. So,

$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

$=2\times \left(2.0\right)+\frac{1}{2}\left(10\right)(2.0{)}^2$

$\text{=24 radian}$

Hence, $\left(D\right)$ is the correct answer.