Question

A particle starting from rest moves with constant tangential acceleration on a circular path of radius $\frac{10}{\pi }\text{m}$. After $2.5$ rotation the velocity of the particle is $50\text{m/s}$, its tangential acceleration is

• A. $25{\text{m/s}}^2$
• B. $25{\text{rad/s}}^2$
• C. $2500{\pi }^2{\text{m/s}}^2$
• D. $2500{\pi }^2{\text{rad/s}}^2$

Answer 1

The correct option is A $25{\text{m/s}}^2$

It is given that the particle strating from rest, its initial angular velocity ${\omega }_0=0$ and let say the final angular velocity is $\omega =\frac{v}{r}=\frac{50\pi }{10}=5\pi \text{rad/s}$

The angular displacement covered after $2.5$ revolution is $\theta =2\pi \times n=2\pi \times 2.5=5\pi \text{rad}$

By the equation of motion in circular kinematics we get,
$\theta =\frac{{\omega }^2-{\omega }_0^2}{2\alpha },\Rightarrow \alpha =\frac{25{\pi }^2}{10\pi }=2.5\pi {\text{rad/s}}^2$

Now we know that the tangential acceleration $a_T=r\alpha$
$\Rightarrow a_T=\frac{10}{\pi }\times 2.5\pi =25{\text{m/s}}^2$

Hence option A is the correct answer