Question 656: Centre of Mass and Rotational Motion

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 point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

Accepted Answer

When the string is cut, the rod will rotate about P. Let
$$\alpha $$ be initial angular acceleration of the rod.Then
Torque, $$	au  = I\alpha  = {{M{L^2}} \over 3}\alpha $$      ...(i)
(Moment of inertia of the rod about one end = $${{M{L^2}} \over 3}$$)
Also, $$	au  = Mg{L \over 2}$$ ...(ii)
Equating (i) and (ii), we get
$$Mg{L \over 2} = {{M{L^2}} \over 3}\alpha  \Rightarrow \alpha  = {{3g} \over {2L}}$$

Answer

$${{2g} \over L}$$

Answer

$${{2g} \over {2L}}$$

Answer

$${{3g} \over {2L}}$$

Answer

$${g \over L}$$

Centre of Mass and Rotational Motion

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 point Q as shown in figure. When string is cut, the initial angular acceleration of the rod is

Choose the correct answer: