Question

A thin uniform rod (\(X\)) of mass \(M\) and length \(L\) is pivoted at a height \( \left(\dfrac{L}{3}\right) \) as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top is ________. (\(g\) = gravitational acceleration) 

• A. \( \sqrt{\dfrac{3g}{2L}} \)
• B. \( \dfrac{3}{\sqrt{2}} \sqrt{\dfrac{g}{L}} \)
• C. \( \sqrt{\dfrac{3g}{L}} \)
• D. \( \dfrac{1}{\sqrt{2}} \sqrt{\dfrac{g}{L}} \)

Hint:

Always use the parallel axis theorem when rotation is about a point other than the centre of mass.

Answer 1

The Correct Option is C

To find the angular velocity of the rod when it hits the table, we need to consider the conservation of energy. Let’s solve the problem step-by-step:

1. Initially, the rod is vertical. The potential energy of the rod with respect to the pivot point is due to its center of mass.
2. The center of mass of the rod is located at a height \(L/2\) from its base.
3. The vertical distance of the center of mass from the pivot when the rod is vertical is \((L/2) - (L/3) = L/6\).
4. The initial potential energy (PE_i) of the rod is given by:

\(PE_i = M g \left(\dfrac{L}{6}\right)\)

1. When the rod falls and is horizontal, all potential energy is converted to rotational kinetic energy (KE_rot).
2. The rotational kinetic energy is given by:

\(KE_{rot} = \dfrac{1}{2} I \omega^2\)

1. Here, \(I\) is the moment of inertia of the rod about the pivot, and \(\omega\) is the angular velocity.
2. The moment of inertia \(I\) for a rod pivoted at a point a distance \(d\) from its center is:

\(I = \dfrac{1}{3} M L^2 + M \left(\dfrac{L}{6}\right)^2\)

\(I = \dfrac{1}{3} M L^2 + \dfrac{M L^2}{36} = \dfrac{13 M L^2}{36}\)

1. Using conservation of energy:

\(M g \left(\dfrac{L}{6}\right) = \dfrac{1}{2} \left(\dfrac{13 M L^2}{36}\right) \omega^2\)

1. Solving for \(\omega\):

\(\omega^2 = \dfrac{6 \times 36 \times g}{13 \times L}\)
\(\omega = \sqrt{\dfrac{216g}{13L}}\)

1. The angular velocity when the rod hits the table is then approximately:

\(\omega = \sqrt{\dfrac{3g}{L}}\)

Thus, the correct answer is:

\(\sqrt{\dfrac{3g}{L}}\)