Rotational Motion

JEE Main level, numerical answer, medium

141 of 280

Question 141 of 280

A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is ____\_\_\_\_ J.(g=10 m/s2)\mathrm{J}.\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)

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Answer: 30

Let the mass of the flywheel be M=3M = 3 kg and its radius be R=5R = 5 m.
Let the hanging mass be m=3m = 3 kg and the distance descended be h=3h = 3 m.
The flywheel is a disk, so its moment of inertia is I=12MR2I = \frac{1}{2}MR^2.
By conservation of energy, the loss in potential energy of the hanging mass equals the gain in kinetic energy of the system.
mgh=12mv2+12Iω2mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
Since the string does not slip, v=Rωv = R\omega, so ω=vR\omega = \frac{v}{R}.
mgh=12mv2+12(12MR2)(vR)2mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2
mgh=12mv2+14Mv2mgh = \frac{1}{2}mv^2 + \frac{1}{4}Mv^2
Substituting the values: 3×10×3=12(3)v2+14(3)v23 \times 10 \times 3 = \frac{1}{2}(3)v^2 + \frac{1}{4}(3)v^2
90=32v2+34v2=94v290 = \frac{3}{2}v^2 + \frac{3}{4}v^2 = \frac{9}{4}v^2
v2=90×49=40v^2 = \frac{90 \times 4}{9} = 40 m2^2/s2^2.
The kinetic energy of the wheel is Kwheel=12Iω2=14Mv2K_{wheel} = \frac{1}{2}I\omega^2 = \frac{1}{4}Mv^2.
Kwheel=14×3×40=30K_{wheel} = \frac{1}{4} \times 3 \times 40 = 30 J.

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