## Physics (10th Edition)

We call the constant angular acceleration $\alpha$ and the angle the drill has moved through $\theta$. It starts from rest, so $\omega_0=0$ The drill's final angular speed is $\omega^2=0^2+2\alpha\theta=2\alpha\theta$ The radius of the rotation circle at the mentioned point is $r$. At that point, we have $$a_c=r\omega^2=2r\alpha\theta$$ $$a_T=r\alpha$$ It is said that $a_c=2a_T$. Therefore, $$2r\alpha\theta=2r\alpha$$ $$\theta=1rad$$