Question

Starting from rest, a particle moves on a circle with constant angular acceleration α. The time at which its tangential and normal acceleration become equal in magnitude is

• A. $\sqrt{\frac{1}{\alpha}}$
• B. $\sqrt{\frac{2}{\alpha}}$
• C. $\sqrt{\frac{3}{\alpha}}$
• D. $\sqrt{\frac{4}{\alpha}}$

Answer 1

Answer: (A)

$\sqrt{\frac{1}{\alpha}}$

Answer 2

Given that the particle starts from rest and has a constant angular acceleration $\alpha$. The tangential acceleration is $a_t = R\alpha$. The angular velocity at time $t$ is $\omega = \alpha t$ (since initial angular velocity is zero). The linear speed is $v = \omega R = (\alpha t)R$. The normal (centripetal) acceleration is $a_n = \frac{v^2}{R} = \frac{(\alpha t R)^2}{R} = \alpha^2 R t^2$. We need to find the time $t$ when the magnitudes of tangential and normal accelerations are equal: $|a_t| = |a_n|$ $R\alpha = \alpha^2 R t^2$ Assuming $R \neq 0$ and $\alpha \neq 0$, we simplify: $1 = \alpha t^2$ Solving for $t^2$: $t^2 = \frac{1}{\alpha}$ Taking the positive square root for time: $t = \sqrt{\frac{1}{\alpha}}$