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JEE Main 2022 (26 Jul, Shift 1) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2022 (26 Jul, Shift 1)

Q33JEE Main 2022 · 26 JUL · SHIFT 1Medium

Rotational Kinematics & Dynamics

A disc of mass 1 kg1\text{ kg} and radius RR is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be 4x3R rad s−14\sqrt{\frac{x}{3R}}\text{ rad s}^{-1} where x=x = ____________. (g=10 m s−2)\left(g = 10\text{ m s}^{-2}\right)

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

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Answer: 5

  1. I=12(1)R2+(1)R2=32R2I = \frac{1}{2}(1)R^2 + (1)R^2 = \frac{3}{2}R^2 (disc + point mass on rim).
  2. Body falls 2R2R: mg(2R)=12Iω2⇒20R=34R2ω2mg(2R) = \frac{1}{2}I\omega^2 \Rightarrow 20R = \frac{3}{4}R^2\omega^2
  3. ω2=803R⇒ω=453R\omega^2 = \frac{80}{3R} \Rightarrow \omega = 4\sqrt{\frac{5}{3R}}
  4. So x=5x = 5.
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