Question

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

Answer 1

Step 1: Given that:

Initial velocity(u) of the particle= $0$

Angular acceleration of the particle is constant and = $\alpha$

Step 2: Calculation of velocity of the particle on the circle:

In circular motion, the centripetal acceleration or normal acceleration of the body is given by,

$a_c=\frac{v^2}{r}$ ,

Where v is the velocity of the body and r is the radius of the circular path.

We know that,

$\omega$=ωo\omega_oωo​+$\alpha$$t$

Since, ${\omega }_o$ = 0

$\omega$=$\alpha$$t$

$\frac{v}{r}=\alpha t$

$v=\alpha tr.......\left(1\right)$

Step 3: Calculation of time:

According to the question,

Normal acceleration= Tangential acceration

$a_c=a_t$

$\frac{v^2}{r}=\alpha r$

From equation (1)

$\alpha r=\frac{{\left(\alpha tr\right)}^2}{r}$

$\therefore t=\frac{1}{\sqrt{\alpha }}$

The time at which its tangential and normal acceleration becomes equal in magnitude is $\sqrt{\frac{1}{\alpha }}$ .