A) \[\frac{2g}{3l}\]
B) \[mg\frac{l}{2}\]
C) \[\frac{3}{2}gl\]
D) \[\frac{3g}{2l}\]
Correct Answer: D
Solution :
The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is \[I=\frac{m{{l}^{2}}}{3}\] where \[m\] is mass of rod and \[l\] is length. Torque \[(\tau =I\alpha )\] acting on centre of gravity of rod is given by \[\tau =mg\frac{l}{2}\] or \[I\alpha =mg\frac{l}{2}\] or \[\frac{m{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\] or \[\alpha =\frac{3g}{2l}\]
You need to login to perform this action.
You will be redirected in
3 sec
