Physics - Rotation Question with Solution | TestHub

Just after the cut, with $\theta$ the angle string $BD$ makes with the vertical: $m{a}_x=-T\sin \theta$, $m{a}_y=T\cos \theta -mg$, and $\frac{m{\ell }^2}{12}\alpha =\frac{T\ell \cos \theta }{2}$ ⇒ $\alpha =\frac{6T\cos \theta }{m\ell }$.

The string stays taut with zero angular speed, so the acceleration of $B$ along $BD$ is zero: $-a_{Bx}\sin \theta +a_{By}\cos \theta =0$.

Substituting: $T(1+3{\cos }^2\theta )=mg\cos \theta$ ⇒ $T=\frac{mg\cos \theta }{3{\cos }^2\theta +1}$ - (A)