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JEE Advanced Physics Rotational Motion 2020 JEE Advanced 2020 (Paper 2)

JEE Advanced Physics Question (2020) — Solution

Question

A rod of mass m and length L , pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it. The combined system now rotates with an angular speed ω about the pivot. The maximum angular speed ω M is achieved for x = x M . Then

Options

  1. A. ω = 3 v x L 2 + 3 x 2
  2. B. ω = 12 v x L 2 + 12 x 2
  3. C. x M = L 3
  4. D. ω M = v 2 L 3

Answer

D. ω M = v 2 L 3

Step-by-step solution

The net torque on the system (rod+bullet) will be zero about hinge point. Therefore, we can apply the principle of angular momentum conservation about hinge point. From angular momentum conservation, m v x = m L 2 3 + m x 2 ω ⇒ ω = 3 v x L 2 + 3 x 2 = 3 v L 2 x + 3 x For maximum angular velocity, d ω d x = 0 ⇒ d d x L 2 x + 3 x = 0   ⇒ - L 2 x 2 + 3 = 0   ⇒ x = L 3 ⇒ ω max = 3 v 3 L 2 L + 3 L = 3 v 2 L

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