Question

A solid body starts rotating about a stationary axis with an angular acceleration $\beta =at$, where $a=0.5\times {10}^{-3}rad/s^2$. If ${}^{\prime }{t}^{\prime }$ sec after the beginning of the rotation, the total acceleration vector of an arbitrary point of the body forms an angle $\alpha ={45}^{\circ }$ with its velocity vector, then $\frac{t}{10}$ is equal to

Answer 1

$\beta =at\Rightarrow \frac{d\omega }{dt}=at$

$\omega =\frac{a{t}^2}{2}$

now

$a_r={\omega }^{2}r=r{a}^2{t}^4/4$

$a_t=r\beta =atr$

Since

$tan{45}^o=\frac{a_r}{a_t}=\frac{a{t}^3}{4}$

$\Rightarrow t=\sqrt[3]{3\frac{4}{a}}=20secs$