91Ó°ÊÓ

The mass of the rod is M.

The length of the rod is l.

The torque,\(\tau \)and angular acceleration,\(\alpha \)are related as:

\(\tau = I\alpha \)

Here, I is the moment of inertia and is given as the product of the mass of the rigid body and the square of its distance from the axis of rotation.

\(I = m{r^2}\)

The angular acceleration can be calculated using the relation between the torque and moment of inertia.

\(\begin{aligned}{c}\tau = I\alpha \\rF = I\alpha \end{aligned}\) … (i)

Here, the torque is provided by gravity and the moment of inertia of the rod about the center of mass is\(I = \frac{1}{3}M{l^2}\). Therefore, from equation (i), you get:

\(\begin{aligned}{c}r\left( {Mg} \right) = \left( {\frac{1}{3}M{l^2}} \right)\alpha \\\alpha = \frac{{3rg}}{{{l^2}}}\\ = \frac{{3\left( {\frac{l}{2}} \right)g}}{{{l^2}}}\\ = \frac{{3g}}{{2l}}\end{aligned}\)

The relation between angular and linear acceleration is:

\(a = \alpha r\)

Here,\(r = l\). Therefore, the linear acceleration of the tip of the rod will be:

\(\begin{aligned}{c}a = \left( {\frac{{3g}}{{2l}}} \right)l\\ = \frac{{3g}}{2}\end{aligned}\)