JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Rotational Mechanics

A thin smooth rod of length

L

and mass

M

is rotating freely with angular speed

ω_{0}

about an axis perpendicular to the rod and passing through its centre. Two beads of mass m and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod will be:-

(A) $\frac{M ω _{0}}{M + 3 m}$
(B) $\frac{M ω _{0}}{M + m}$
(C) $\frac{M ω _{0}}{M + 2 m}$
(D) $\frac{M ω _{0}}{M + 6 m}$ ✓ Correct

Answer: (D)

Solution

There is no external torque about the rotation axis, so the angular momentum of the rod-plus-beads system stays fixed. With the beads at the centre they add nothing, so initially

L = \frac{M L ^{2}}{12} ω_{0}

. When the beads slide to the two ends, each sits at

\frac{L}{2}

and contributes

m (\frac{L}{2})^{2}

, so

L = [\frac{M L ^{2}}{12} + 2 m \frac{L ^{2}}{4}] ω_{f} = \frac{L ^{2}}{12} (M + 6 m) ω_{f}

. Equating the two expressions,

ω_{f} = \frac{M ω _{0}}{M + 6 m}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Rotational Mechanics →

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Rotational Mechanics

Solution

More JEE Main questions on Rotational Mechanics