JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Rotational Mechanics

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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 a 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 / s^{2}

)

Answer: 30

Solution

As the

3

kg block descends by

h = 3

m, its lost potential energy becomes the kinetic energy of the falling block and of the rotating flywheel. Taking the flywheel as a disc,

I = \frac{1}{2} M R^{2}

, and the string imposes

v = ω R

. Energy conservation gives

m g h = \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2} = \frac{1}{4} M v^{2} + \frac{1}{2} m v^{2}

. With

M = m = 3

kg,

m g h = \frac{3}{4} m v^{2}

, so

v^{2} = \frac{4 g h}{3} = \frac{4 \times 10 \times 3}{3} = 40 m^{2} / s^{2}

. The kinetic energy of the flywheel is

\frac{1}{2} I ω^{2} = \frac{1}{4} M v^{2} = \frac{1}{4} \times 3 \times 40 = 30

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Rotational Mechanics

Solution

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