Question 550: Centre of Mass and Rotational Motion

Question

A thin uniform $$\operatorname{rod} A B$$ of mass $$m$$ and length $$l$$ is hinged at one end $$A$$ to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its end $$B$$ strikes the ground is ( $$g=$$ acceleration due to gravity)

Accepted Answer

Let the length of the rod $$A B$$ be $$l$$ and mass of the rod be $$m$$.
Moment of inertia of the uniform rod about an axis passing through one of its end $$V$$ given as,
$$\mathrm{MI}=\frac{m l^2}{3}$$
Rotational kinetic energy, $$(\mathrm{KE})_{\mathrm{rod}}=\frac{1}{2} l \omega^2$$
Initially the rod is held, vertical, so potential energy of rod is given by $$(\mathrm{PE})_i=m g \frac{I}{2}\quad 
\left(\because \frac{l}{2}=\mathrm{COM}ight)$$
Initial kinetic energy, $$(\mathrm{KE})_i=0$$
The rod is allowed to fall, when it just touches the ground its potential energy becomes zero, i.e., $$(\mathrm{PE})_f=0$$
Let, angular velocity of rod at that instant be $$\omega$$, then kinetic energy at that instant is
$$(\mathrm{KE})_f=\frac{1}{2} / \omega^2$$
From conservation of energy,
$$(\mathrm{KE})_i+(\mathrm{PE})_i=(\mathrm{KE})_f+(\mathrm{PE})_f$$
$$\begin{array}{rlrl}
\Rightarrow & 0+m g \frac{l}{2} =\frac{1}{2} l \omega^2+0 \
\Rightarrow & m g l =l \omega^2 \
\Rightarrow & m g l =\frac{m l^2}{3} \omega^2 \
\Rightarrow & \omega =\sqrt{\frac{3 g}{l}}
\end{array}$$

Answer

$$\sqrt{\frac{g}{l}}$$

Answer

$$\sqrt{\frac{m g}{l}}$$

Answer

$$\sqrt{\frac{3 g}{l}}$$

Answer

$$\sqrt{\frac{m g}{3 l}}$$

Centre of Mass and Rotational Motion

Choose the correct answer: