Rotational Mechanics! (Basic)

Firstly, we need to understand how the acceleration will ever be at an angle to the velocity. Whenever a body is in circular motion, it feels a centrifugal acceleration away from the axis of rotation. Moreover, in our case, there is also an angular acceleration and hence a tangential linear acceleration.

These two perpendicular accelerations will form a resultant that would be at an angle to the tangential acceleration, or more importantly to the direction of velocity.

These are the directions!

So, let's consider the radius of the circular path to be $r$ , angular velocity as $\omega$ and tangential acceleration $a$ that can be given by the relation $a = r \alpha$ .

Also, the centrifugal acceleration has the same magnitude as the centripetal acceleration and can be given by ${a}_{centrifugal} = r {\omega}^{2}$ . Now, drawing the free body diagram and by a little geometry, we can see the following picture.

In the above shown diagram, we can see that $\tan{60} = \frac {OB}{OA}$ , or

$\sqrt{3} = \frac {r {\omega}^{2}} {r \alpha}$

But, we know that $\omega = \int { \alpha } dt$ . Hence, from the given information, $\omega = {10}^{-2} {t}^{2} rad {s}^{-1}$

Now, putting these values in the formed equation, we get ${10}^{-4} {t}^{4} = \sqrt{3} (2 \times {10}^{-2}) t$ , or

$t = {(2 \sqrt { 3 } \times {10}^{2})}^{\frac {1}{3}} = 7.023$