Question

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first $2\text{sec}$ it rotates through an angle ${\theta }_1$, in the next $2\text{sec}$ it rotates through an additional angle ${\theta }_2$. The ratio of $\frac{{\theta }_2}{{\theta }_1}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

The correct option is C $3$

Let the uniform angular acceleration be $\alpha$, the initial angular velocity ${\omega }_0=0$

We know that, the angular displacement for first two sec,
${\theta }_1={\omega }_{0}t+\frac{1}{2}\alpha t^2$, $t=2\text{s}$
${\theta }_1=2\alpha$

let the angular velocity changes from ${\omega }_0$ to ${\omega }^{\prime }$ for the first $2\text{s}$, by the equation of motion under constant angular acceleration
${\omega }^{\prime }={\omega }_0+\alpha t\Rightarrow {\omega }^{\prime }=0+2\alpha =2\alpha$

Now for the angular displacement for next time interval of $2\text{s}$
${\theta }_2={\omega }^{\prime }t+\frac{1}{2}\alpha t^2$, $t=2\text{s}$
$\Rightarrow {\theta }_2=4\alpha +2\alpha =6\alpha$

Hence the ratio,

$\frac{{\theta }_2}{{\theta }_1}=\frac{6\alpha }{2\alpha }=3$

(OR)

Let the uniform angular acceleration be $\alpha$, the initial angular velocity ${\omega }_0=0$

The angular displacement for first two sec,
${\theta }_1={\omega }_{0}t+\frac{1}{2}\alpha t^2$, $t=2\text{s}$
${\theta }_1=2\alpha$
The angular displacement for first four sec
${\theta }_1+{\theta }_2={\omega }_{0}t+\frac{1}{2}\alpha t^2$ , $t=4$
${\theta }_1+{\theta }_2=8\alpha$
${\theta }_2=6\alpha$
Hence the ratio,

$\frac{{\theta }_2}{{\theta }_1}=\frac{6\alpha }{2\alpha }=3$

Option C is the correct answer.