Question

A particle is revolving with a constant angular acceleration $\alpha$ in a circular path of radius r. Find the time when the centripetal acceleration will be equal to the tangential acceleration:

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

Answer 1

The correct option is B $\frac{1}{\sqrt{\alpha }}$

Given that,

Angular acceleration $=\alpha$

Now, centripetal acceleration $a=\frac{v^2}{r}$

To start with condition of movement

$\omega =\omega 0+\alpha t$

Now,

$\omega 0=0$

So,

$\omega =\alpha t$

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

$v=r\alpha t$

Now, the centripetal acceleration is equal to the tangential acceleration

$\alpha r=a$

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

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

$t^2=\frac{1}{\alpha }$

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

Hence, the time is $\frac{1}{\sqrt{\alpha }}$