JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Rotational Mechanics

As shown in the figure, a bob of mass

m

is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius

r

and mass

m

. When released from rest the bob starts falling vertically. When it has covered a distance of

h

, the angular speed of the wheel will be

Answer: (D)

Solution

As the bob of mass

m

falls a distance

h

, its lost potential energy is shared between its own translation and the disc's rotation,

m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}

. For the disc

I = \frac{m r ^{2}}{2}

, and the string unwinds without slipping so

v = ω r

. Substituting,

m g h = \frac{1}{2} m ω^{2} r^{2} + \frac{1}{4} m ω^{2} r^{2} = \frac{3}{4} m ω^{2} r^{2}

, hence

ω = \frac{1}{r} \frac{4 g h}{3}

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option