A disc of mass 1 kg and radius R is free of rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body…
JEE Main 2022 — Physics Mechanics
2022integerhard
A disc of mass 1kg and radius R is free of 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 43Rxrads−1 where x= _____ .
Official previous-year question
Held on 26 Jul 2022 · Verified 6 Jul 2026.
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Solution
Let the angular speed of disc be ω.
Using conservation of mechanical energy
mg2R=21Idiscω2+21Iparticleω2
Where, I is moment of inertia.
⇒mg2R=2ω2[2m2+mR2]
⇒mg2R=2ω2(23mR2)
⇒43ω2=R2g
⇒ω2=3R8g
Thus, angular speed is ω=3R80rads−1
Given ω=43Rx
Comparing both, 163Rx=3R80
Therefore, the value of x=5.
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