A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass 0.1 kg moving in a straight line with velocity $80 \mathrm{~m} / \mathrm{s}$ hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in $\mathrm{rad} / \mathrm{s}$) of the rod immediately after the collision will be ____.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Conservation of Momentum

A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass 0.1 kg moving in a straight line with velocity

80 m / s

hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in

rad / s

) of the rod immediately after the collision will be ____.

Answer: 20

Solution

Angular momentum about the pivot is conserved during the collision. Just before impact the particle carries

L_{i} = m v L = 0.1 \times 80 \times 1 = 8

kg m

^{2}

^{- 1}

about the top end. Afterwards the rod and the embedded particle turn together with moment of inertia

I = \frac{M L ^{2}}{3} + m L^{2} = \frac{0.9}{3} + 0.1 = 0.4

kg m

^{2}

. Setting

m v L = I ω

8 = 0.4 ω

, so the angular speed immediately after the collision is

ω = 20

rad/s.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Integer answer

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Conservation of Momentum

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