Question

For a particle moving in a circle of radius $1\text{m}$, speed increases uniformly from $5\text{m/s}$ to $10\text{m/s}$ in $10\text{seconds}$. The value of angular acceleration of the particle is

• A. $50{\text{rad/s}}^2$
• B. $5{\text{rad/s}}^2$
• C. $0.5{\text{rad/s}}^2$
• D. $0.05{\text{rad/s}}^2$

Answer 1

The correct option is C $0.5{\text{rad/s}}^2$

Given, radius of the circle $r=1\text{m}$
initial speed of the particle $u=5\text{m/s}$
final speed of the particle $v=10\text{m/s}$
Since, angular speed is given as $\omega =\frac{\text{Linear Speed}}{\text{radius}}$
So, we have
initial angular speed ${\omega }_i=5\text{rad/s}$
final angular speed ${\omega }_f=10\text{rad/s}$
Time taken by the particle in such change is given as $t=10\text{sec}$
Thus, we have angular acceleration as
${\omega }_f={\omega }_i+\alpha t$
$\Rightarrow \alpha =\frac{{\omega }_f-{\omega }_i}{t}=\frac{10-5}{10}=0.5{\text{rad/s}}^2$