Question

A uniform rod of mass 2m and length L with a small ball of mass m attached at one end is released from horizontal position as shown in figure. Other end of the rod is hinged. Angular speed of rod when it becomes vertical is

• A. $\sqrt{\frac{5g}{3L}}$
• B. $\sqrt{\frac{3g}{L}}$
• C. $\sqrt{\frac{12g}{5L}}$
• D. $\sqrt{\frac{3g}{2L}}$

Answer 1

Answer: (C)

$\sqrt{\frac{12g}{5L}}$

Answer 2

Initial gravitational potential energy lost = Energy gained in rotation.

For the rod:

Mass = 2m, center of mass drops by L/2.

ΔPE(rod) = 2m * g * (L/2) = m * g * L.

For the ball:

Mass = m, drop = L.

ΔPE(ball) = m * g * L.

Total ΔPE = m * g * L + m * g * L = 2m * g * L.

Moment of inertia about the hinge:

Rod: I(rod) = (1/3)(2m)L² = (2mL²)/3.

Ball: I(ball) = mL².

Total I = (2mL²)/3 + mL² = (5mL²)/3.

Energy conservation:

(1/2) * I * ω² = 2m * g * L

⇒ (1/2) * (5mL²/3) * ω² = 2m * g * L

Simplify: (5mL²ω²)/(6) = 2m * g * L

Cancel m and one L: (5Lω²)/(6) = 2g

⇒ ω² = (12g)/(5L)

⇒ ω = √(12g/(5L)).