Rotational Motion

JEE Main level, numerical answer, medium

14 of 280

Question 14 of 280

Two masses mm and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius r=0.1 mr=0.1 \mathrm{~m}. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m mass is released from rest and its speed when it has descended through a height of 3.6 m is ____\_\_\_\_ m/s\mathrm{m} / \mathrm{s}. (Assume string does not slip and g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2})

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Answer: 2

Let the speed of the masses be vv and the angular velocity of the pulley be ω=v/r\omega = v/r since the string does not slip.
The pulley is a disc of mass M=30mM = 30m and radius rr. Its moment of inertia is I=12Mr2=12(30m)r2=15mr2I = \frac{1}{2}Mr^2 = \frac{1}{2}(30m)r^2 = 15mr^2.
Using the principle of conservation of mechanical energy, the loss in potential energy of the system equals the gain in kinetic energy.
When the mass 2m2m descends by h=3.6h = 3.6 m, the mass mm ascends by h=3.6h = 3.6 m.
Loss in potential energy ΔU=(2m)gh−(m)gh=mgh\Delta U = (2m)gh - (m)gh = mgh.
Gain in kinetic energy ΔK=12(2m)v2+12(m)v2+12Iω2\Delta K = \frac{1}{2}(2m)v^2 + \frac{1}{2}(m)v^2 + \frac{1}{2}I\omega^2.
Substituting II and ω\omega: ΔK=32mv2+12(15mr2)(vr)2=32mv2+152mv2=9mv2\Delta K = \frac{3}{2}mv^2 + \frac{1}{2}(15mr^2)(\frac{v}{r})^2 = \frac{3}{2}mv^2 + \frac{15}{2}mv^2 = 9mv^2.
Equating ΔU=ΔK\Delta U = \Delta K: mgh=9mv2mgh = 9mv^2.
v2=gh9v^2 = \frac{gh}{9}.
Given g=10g = 10 m/s2^2 and h=3.6h = 3.6 m:
v2=10×3.69=369=4v^2 = \frac{10 \times 3.6}{9} = \frac{36}{9} = 4.
v=4=2v = \sqrt{4} = 2 m/s.

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