Question

A uniform rod of length 60 cm is placed with one end in contact with the horizontal table and is then inclined at an angle 30\(^\circ\) to the horizontal and allowed to fall. The angular velocity of the rod when it becomes horizontal is (acceleration due to gravity = 10 ms\(^{-2}\))

• A. 9 rad s\(^{-1}\)
• B. 6 rad s\(^{-1}\)
• C. 5 rad s\(^{-1}\)
• D. 8 rad s\(^{-1}\)

Hint:

Use conservation of mechanical energy: Loss in PE = Gain in KE.
PE of CM = \(mgh_{CM}\).
Rotational KE = \(\frac{1}{2}I\omega^2\).
Moment of inertia of a uniform rod about one end = \(\frac{1}{3}mL^2\).
Ensure units are consistent (e.g., meters for length if g is in m/s\(^2\)).

Answer 1

The Correct Option is C

Let the length of the rod be \(L = 60 \text{ cm} = 0.6 \text{ m}\). Mass of the rod be \(m\). The rod is allowed to fall, with one end in contact with the horizontal table. This end acts as a pivot. Initially, the rod is inclined at \(\theta_0 = 30^\circ\) to the horizontal. The height of the center of mass (CM) initially is \(h_1 = (L/2)\sin\theta_0\). When the rod becomes horizontal, its CM is at height \(h_2 = 0\) (if the pivot is on the table level) or very close to the table surface if it's a thin rod. Let's assume the pivot end is fixed on the table, so when horizontal, the CM is at height 0 relative to its lowest possible point if it were swinging through. More precisely, if one end is on the table and it falls to be horizontal, the CM's height changes from \( (L/2)\sin 30^\circ \) to 0 (relative to the table height if CM could go below, but it becomes horizontal on the table). This is a conservation of energy problem. Loss in Potential Energy (PE) = Gain in Rotational Kinetic Energy (KE). Initial PE (of CM, taking table level as reference for final state of CM being at some height if rod has thickness, but for a thin rod, let's assume CM height becomes 0). If the rod falls such that it lies flat on the table, the CM height drops from \( (L/2)\sin 30^\circ \) to essentially 0 (or a very small height if rod has thickness). Initial height of CM above the table surface: \(h_i = (L/2)\sin 30^\circ\). Final height of CM when rod is horizontal (assuming pivot at table level, rod lies flat): \(h_f = 0\). Loss in PE = \(mg(h_i - h_f) = mg(L/2)\sin 30^\circ\). Gain in Rotational KE = \(\frac{1}{2}I\omega^2\), where I is the moment of inertia about the pivot (the end in contact with the table). Moment of inertia of a uniform rod of mass m and length L about one end is \(I = \frac{1}{3}mL^2\). So, \(mg(L/2)\sin 30^\circ = \frac{1}{2} (\frac{1}{3}mL^2) \omega^2\). \(g(L/2)(1/2) = \frac{1}{6}L^2 \omega^2\) (mass m cancels). \( \frac{gL}{4} = \frac{L^2 \omega^2}{6} \) \( \omega^2 = \frac{6gL}{4L^2} = \frac{3g}{2L} \) \( \omega = \sqrt{\frac{3g}{2L}} \). Given \(L=0.6\) m, \(g=10 \text{ ms}^{-2}\). \( \omega = \sqrt{\frac{3 \times 10}{2 \times 0.6}} = \sqrt{\frac{30}{1.2}} = \sqrt{\frac{300}{12}} = \sqrt{25} = 5 \text{ rad s}^{-1}\). This matches option (c). \[ \boxed{5 \text{ rad s}^{-1}} \]