Question

A wheel has angular acceleration of $3.0rad/s^2$ and an initial angular speed of $2.00rad/s$. In a time of $2s$ it has rotated through an angle (in radian) of

• A. 6
• B. 10
• C. 12
• D. 4

Answer 1

Answer: (B)

Angular acceleration is time derivative of angular speed and angular speed is time derivative of angular displacement.
By definition $\alpha =\frac{d\omega }{dt}$
ie, $d\omega =\alpha dt$
So, if in time $t$ the angular speed of a body changes from ${\omega }_0$ to $\omega$
$\underset{{\omega }_0}{\overset{\omega }{\int }}d\omega =\underset{0}{\overset{t}{\int }}\alpha dt$
If $\alpha$ is constant
$\omega -{\omega }_0=\alpha t$
or $\omega ={\omega }_0+\alpha t$ ... (i)
Now, as by definition
$\omega =\frac{d\theta }{dt}$
Eq. (i) becomes
$\frac{d\theta }{dt}={\omega }_0+\alpha t$
$d\theta =\left({\omega }_0+\alpha t\right)dt$
So, if in the time $t$ angular displacement is $\theta$.
$\underset{0}{\overset{\theta }{\int }}d\theta =\underset{0}{\overset{t}{\int }}\left({\omega }_0+\alpha t\right)dt$
or $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$ ... (ii)
Given, $\alpha =3.0rad/s^2$,
${\omega }_0=2.0rad/s,t=2s$
Hence, $\theta =2\times 2+\frac{1}{2}\times 3\times (2{)}^2$
or $\theta =4+6=10rad$
Note Eqs. (i) and (ii) are similar to first and second equations of linear motion.