Question
A rod $$PQ$$ of mass $$M$$ and length $$L$$ is hinged at end $$P.$$ The rod is kept horizontal by a massless string tied to a point $$Q$$ as shown in figure. When string is cut, the initial angular acceleration of the rod is
A rod $$PQ$$ of mass $$M$$ and length $$L$$ is hinged at end $$P.$$ The rod is kept horizontal by a massless string tied to a point $$Q$$ as shown in figure. When string is cut, the initial angular acceleration of the rod is
A.
$$\frac{{3g}}{{2L}}$$
B.
$$\frac{g}{L}$$
C.
$$\frac{{2g}}{L}$$
D.
$$\frac{{2g}}{{3L}}$$
Answer :
$$\frac{{3g}}{{2L}}$$
Solution :
Torque on the rod is equal to moment of weight of rod about $$P.$$
Torque on the rod = Moment of weight of the rod about $$P$$
$$\tau = Mg\frac{L}{2}\,......\left( {\text{i}} \right)$$
$$\because $$ Moment of inertia of rod
about $$P = \frac{{M{L^2}}}{3}\,......\left( {{\text{ii}}} \right)$$
As $$\tau = I\alpha $$
From Eqs. (i) and (ii), we get
$$Mg\frac{L}{2} = \frac{{M{L^2}}}{3}\alpha \Rightarrow \alpha = \frac{{3g}}{{2L}}$$
Torque on the rod is equal to moment of weight of rod about $$P.$$
Torque on the rod = Moment of weight of the rod about $$P$$
$$\tau = Mg\frac{L}{2}\,......\left( {\text{i}} \right)$$
$$\because $$ Moment of inertia of rod
about $$P = \frac{{M{L^2}}}{3}\,......\left( {{\text{ii}}} \right)$$
As $$\tau = I\alpha $$
From Eqs. (i) and (ii), we get
$$Mg\frac{L}{2} = \frac{{M{L^2}}}{3}\alpha \Rightarrow \alpha = \frac{{3g}}{{2L}}$$
