JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Rotational Mechanics

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 angular speed

ω

about the pivot. The maximum angular speed

ω_{M}

is achieved for

x = x_{M}

. Then

Answer: (A), (C), (D)

Solution

By conservation of angular momentum about the pivot:

m v x = I ω

where

I = \frac{m L ^{2}}{3} + m x^{2}

(moment of inertia of rod + bullet)

m v x = (\frac{m L ^{2}}{3} + m x^{2}) ω

ω = \frac{v x}{\frac{L ^{2}}{3} + x ^{2}} = \frac{3 v x}{L ^{2} + 3 x ^{2}}

✓ (Option A)

For maximum

ω

\frac{d ω}{d x} = 0

\frac{d}{d x} (\frac{3 v x}{L ^{2} + 3 x ^{2}}) = 0

3 v (L^{2} + 3 x^{2}) - 3 v x (6 x) = 0

L^{2} - 3 x^{2} = 0

x_{M} = \frac{L}{3}

✓ (Option C)

Substituting

x_{M}

into the expression for

ω

ω_{M} = \frac{3 v \cdot \frac{L}{3}}{L ^{2} + 3 \cdot \frac{L ^{2}}{3}} = \frac{3 v L}{2 L ^{2}} = \frac{v 3}{2 L}

✓ (Option D)

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options