Question

A uniform rod AB of length 1 m and mass 4 kg is sliding along two mutually perpendicular frictionless walls OX and OY. The velocity of the two ends of the rod A and Bare 3 m/s and 4 m/s respectively, as shown in the figure. Then which of the following statement(s) is/are correct?

 

• A. The velocity of the centre of mass of the rod is 2.5 m/s.
• B. Rotational kinetic energy of the rod is \(\frac{25}{6}\) joule.
• C. The angular velocity of the rod is 5 rad/s clockwise.
• D. The angular velocity of the rod is 5 rad/s anticlockwise.

Hint:

The velocity of the centre of mass of an object can be found by averaging the velocities of the points of the object weighted by their masses (for a uniform rod, mass is evenly distributed).

Answer 1

The Correct Option is A, B, D

1. Step 1: To calculate the center of mass velocity, the formula is:

   \( v_{\text{cm}} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \)

   Where \( v_1 \) and \( v_2 \) are end velocities, and \( m_1 \) and \( m_2 \) are their masses. With a uniform rod, \( m_1 = m_2 = \frac{m}{2} \).

2. Step 2: Given end velocities \( v_1 = 3 \, \text{m/s} \) and \( v_2 = 4 \, \text{m/s} \), substitute into the formula:

   \( v_{\text{cm}} = \frac{1}{2} \times (v_1 + v_2) = \frac{1}{2} \times (3 + 4) = 2.5 \, \text{m/s} \)

   Therefore, the center of mass velocity is \( v_{\text{cm}} = 2.5 \, \text{m/s} \).

3. Step 3: Rotational kinetic energy and angular velocity involve rod rotational dynamics. However, the primary result is the center of mass velocity.

Conclusion: The center of mass velocity is \( v_{\text{cm}} = 2.5 \, \text{m/s} \).