Question

A uniform rod of length \(60\ \text{cm}\) 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
\[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]

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

Hint:

For a uniform rod rotating about one end, \[ I=\frac{1}{3}mL^2. \] When a rod falls under gravity, use conservation of energy: \[ \text{Loss in P.E.}=\text{Gain in rotational K.E.} \] to find the angular velocity.

Answer 1

The Correct Option is C

Step 1: Identify the axis of rotation.
One end of the rod remains in contact with the horizontal table and acts as the pivot.
Length of the rod is \[ L=60\ \text{cm}=0.6\ \text{m}. \] Initially, the rod is inclined at \[ 30^\circ \] to the horizontal and is released from rest.

Step 2: Find the loss in gravitational potential energy.
The centre of mass of a uniform rod is at its midpoint.
Initial height of the centre of mass above the table is \[ h_i=\frac{L}{2}\sin30^\circ. \] \[ h_i=\frac{0.6}{2}\times \frac{1}{2}. \] \[ h_i=0.15\ \text{m}. \] When the rod becomes horizontal, the centre of mass lies on the horizontal level of the pivot, so \[ h_f=0. \] Hence, the decrease in potential energy is \[ \Delta U=mg(h_i-h_f). \] \[ \Delta U=mg(0.15). \]

Step 3: Convert the lost potential energy into rotational kinetic energy.
Since the rod rotates about one end, \[ I=\frac{1}{3}mL^2. \] The rotational kinetic energy is \[ K=\frac{1}{2}I\omega^2. \] Using conservation of mechanical energy, \[ mg(0.15) = \frac{1}{2}\left(\frac{1}{3}mL^2\right)\omega^2. \] Substituting \[ g=10,\qquad L=0.6, \] we get \[ m(10)(0.15) = \frac{1}{6}m(0.6)^2\omega^2. \] \[ 1.5m = \frac{1}{6}(0.36)m\omega^2. \] \[ 1.5 = 0.06\omega^2. \]

Step 4: Calculate the angular velocity.
\[ \omega^2=\frac{1.5}{0.06}. \] \[ \omega^2=25. \] Therefore, \[ \omega=5\ \text{rad s}^{-1}. \]

Step 5: Final conclusion.
Hence, the angular velocity of the rod when it becomes horizontal is \[ \boxed{5\ \text{rad s}^{-1}} \] Therefore, the correct option is \[ \boxed{(3)} \]