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JEE Main 2026 (28 Jan, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2026 (28 Jan, Shift 2)

Q7JEE Main 2026 · 28 JAN · SHIFT 2Medium

Rotational Kinematics & Dynamics

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/s2g = 10\text{ m/s}^{2})

Numerical answer type. Work it out, then open the answer.

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Answer: 30

  1. Treat the flywheel as a disc: KEwheel=12⋅12MR2ω2=14Mv2KE_{wheel} = \frac{1}{2}\cdot\frac{1}{2}MR^2\omega^2 = \frac{1}{4}Mv^2, KEmass=12mv2KE_{mass} = \frac{1}{2}mv^2.
  2. mgh=3×10×3=90mgh = 3\times10\times3 = 90 J =(34+32)v2=94v2⇒v2=40= \left(\frac{3}{4} + \frac{3}{2}\right)v^2 = \frac{9}{4}v^2 \Rightarrow v^2 = 40
  3. KEwheel=14(3)(40)=30KE_{wheel} = \frac{1}{4}(3)(40) = 30 J
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