Consider a uniform rod of mass $\mathrm{M}=4 \mathrm{~m}$ and length $\ell$ pivoted about its centre. A mass m moving with velocity v making angle $\theta=\frac{\pi}{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular sped of the rod-mass system just after the collision is

(B) $\frac{3 \sqrt{2}}{7} \frac{v}{\ell}$

(A) $\frac{3}{7} \frac{\mathrm{v}}{\ell}$

(C) $\frac{4}{7} \frac{\mathrm{v}}{\ell}$

(D) $\frac{3}{7 \sqrt{2}} \frac{v}{\ell}$

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Rotational Mechanics

Consider a uniform rod of mass

M = 4 m

and length

ℓ

pivoted about its centre. A mass m moving with velocity v making angle

θ = \frac{π}{4}

to the rod's long axis collides with one end of the rod and sticks to it. The angular sped of the rod-mass system just after the collision is

Answer: (B)

Angular momentum about the central pivot is conserved in the collision. The incoming mass

m

has velocity component

v sin (π /4) = v / 2

perpendicular to the rod, acting at a distance

ℓ /2

from the centre, so

L_{i} = m (v / 2) (ℓ /2)

. After it sticks the moment of inertia is

I_{f} = \frac{M ℓ ^{2}}{12} + m (\frac{ℓ}{2})^{2} = \frac{4 m ℓ ^{2}}{12} + \frac{m ℓ ^{2}}{4} = \frac{7 m ℓ ^{2}}{12}

. Setting

L_{i} = I_{f} ω

gives

\frac{m v ℓ}{2 2} = \frac{7 m ℓ ^{2}}{12} ω

, so

ω = \frac{6 v}{7 2 ℓ} = \frac{3 2}{7} \frac{v}{ℓ}

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option