Question

A solid body starts rotating about a stationary axis with an angular acceleration $\alpha =(2.0\times {10}^{-2})trad/s^2$ , where $t$ is in seconds. How soon(in seconds) after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle $\theta ={60}^{\circ }$ with its velocity vector?

Answer 1

$tan{60}^{\circ }=\frac{a_r}{a_t}=\frac{r{\omega }^2}{r\alpha }$
$=\frac{{\left[{\int }_0^t\alpha dt\right]}^2}{\alpha }$
$=\frac{(0.01t^2{)}^2}{0.02t}$
Solving we get,
$t^3=346$
$\therefore t=7s$