A) \[\frac{3g}{2\ell }\cos \theta \]
B) \[\frac{2g}{3\ell }\cos \theta \]
C) \[\frac{3g}{2\ell }\sin \theta \]
D) \[\frac{2g}{3\ell }\sin \theta \]
Correct Answer: C
Solution :
Taking torque about pivot \[\tau =I\alpha \] \[mg\sin \theta \frac{\ell }{2}=\frac{m{{\ell }^{2}}}{3}\alpha \] \[\alpha =\frac{3g}{2\ell }\sin \theta \]You need to login to perform this action.
You will be redirected in
3 sec