Question

A thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega $. Its centre of mass rises to a maximum height of:
(A) $\dfrac{{{l^2}{\omega ^2}}}{{3g}}$
(B) $\dfrac{{{l^2}{\omega ^2}}}{{8g}}$
(C) $\dfrac{{{l^2}{\omega ^2}}}{{2g}}$
(D) $\dfrac{{{l^2}{\omega ^2}}}{{6g}}$

Answer 1

Hint: To solve this question, we need to use the energy conservation principle. The total mechanical energy will be conserved for the two instants given in the question.

Formula used:
$\Rightarrow K = \dfrac{1}{2}I{\omega ^2}$, where $K$ is the kinetic energy of rotation of a body about an axis having the angular velocity of $\omega $, and moment of inertia $I$ about the same axis.

Complete step by step solution:
Let the maximum height reached by the centre of mass of the rod be $h$
We know that the kinetic energy of the rotation of a body is given by
$\Rightarrow K = \dfrac{1}{2}I{\omega ^2}$
When the centre of mass of the rod is at the lowest position, the rod will have minimum potential energy and maximum kinetic energy. From the above expression, we can say that at this position, the rod will have maximum angular velocity. We take the potential energy of the rod at this point to be zero.
So, the total energy at this point is equal to the kinetic energy, that is
$\Rightarrow {E_1} = \dfrac{1}{2}I{\omega ^2} + 0$
$\Rightarrow {E_1} = \dfrac{1}{2}I{\omega ^2}$ …………………..(i)
Now at the point when the centre of mass of the rod is at the maximum height, the rod will have zero kinetic energy. So its total energy at this point is equal to the potential energy
$\Rightarrow {E_2} = 0 + mgh$
$\Rightarrow {E_2} = mgh$ ………………...(ii)
As there is no non-conservative force acting on the rod, so its total mechanical energy will remain constant. That is,
$\Rightarrow {E_1} = {E_2}$
From (i) and (ii)
$\Rightarrow \dfrac{1}{2}I{\omega ^2} = mgh$
$\Rightarrow h = \dfrac{1}{2}\dfrac{I}{{mg}}{\omega ^2}$ ………………...(iii)
Now, we need to find the moment of inertia $I$of the rod about the given axis.
We know that the moment of inertia of a uniform rod about an axis passing through its centre is
$\Rightarrow \dfrac{{m{l^2}}}{{12}}$
Now, in this case the axis is passing through its end. So we apply the parallel axis theorem to get the moment of inertia as
$\Rightarrow I = \dfrac{{m{l^2}}}{{12}} + m{\left( {\dfrac{l}{2}} \right)^2}$
$\Rightarrow I = \dfrac{{m{l^2}}}{{12}} + \dfrac{{m{l^2}}}{4}$
On solving we get
$\Rightarrow I = \dfrac{{m{l^2}}}{3}$
Putting this in (iii)
$\Rightarrow h = \dfrac{1}{6}\dfrac{{m{l^2}}}{{mg}}{\omega ^2}$
Cancelling$m$we finally get
$\Rightarrow h = \dfrac{{{\omega ^2}{l^2}}}{{6g}}$
Thus the maximum height that the centre of mass rises to is equal to $\dfrac{{{\omega ^2}{l^2}}}{{6g}}$.

Hence, the correct answer is option (D).

Note:
If we do not remember the formula for the kinetic energy of rotation, we can easily derive it using the analogy between translational and the rotational motion. We know the formula for the kinetic energy in translational motion. Replacing the translational variables with the analogous rotational variables will give the expression for the kinetic energy of rotation.