Chapter 10 - Rotational Kinematics and Energy - Problems and Conceptual Exercises - Page 326: 45

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

We know that the tangential acceleration is given as $a_t=R\alpha$ and the centripetal acceleration is given as $a_{cp}=R\omega^2$ $a_{cp}=R(\alpha t)^2$ $a_{cp}=R\alpha^2t^2$ Given that $a_t=a_{cp}$ $\implies R\alpha=R\alpha^2t^2$ $\implies I=\alpha t^2$ This can be rearranged as: $t^2=\frac{1}{\alpha}$ $\implies t=\sqrt{\frac{1}{\alpha}}$