Question

A particle starting from rest, moves in a circle of radius 'r'. It attains a velocity of V0 m/s in the $n^{th}$ round. Its angular acceleration will be :-

• A. $\frac{\vee_{0}}{n}rad / s^{2}$
• B. $\frac{\vee^{2}_{0}}{2\pi nr^{2}}rad / s^{2}$
• C. $\frac{\vee^{2}_{0}}{4\pi nr^{2}}rad / s^{2}$
• D. $\frac{\vee^{4}_{0}}{4\pi nr}rad / s^{2}$

Answer 1

The Correct Option is C

To find the angular acceleration of a particle moving in a circle of radius r, starting from rest and attaining a velocity V_0 m/s in the n^{th} round, we need to consider the relationship between linear and angular quantities.

The relationship between linear velocity v and angular velocity \omega is given by:

v = \omega \cdot r

Given that the particle attains a velocity of V_0 m/s, the corresponding angular velocity \omega_0 when it completes the n^{th} round is:

\omega_0 = \frac{V_0}{r}

The number of radians in one complete circle is 2\pi. Therefore, for n rounds, the total angular displacement \theta is:

\theta = 2\pi n

Since the particle starts from rest, initial angular velocity \omega_1 = 0. The angular acceleration \alpha can be calculated using the kinematic equation:

\omega_0^2 = \omega_1^2 + 2\alpha \theta

Substituting the known values, we have:

\left(\frac{V_0}{r}\right)^2 = 0 + 2\alpha \cdot 2\pi n

Solving for \alpha:

\frac{V_0^2}{r^2} = 4\pi n \alpha

\alpha = \frac{V_0^2}{4\pi n r^2}

This matches the correct answer from the given options, confirming that the angular acceleration is:

\frac{V_0^2}{4\pi nr^2} \text{ rad/s}^2