Torque of centrifugal force τcf=dm.xsinθω2xcosϑ=lmω2sinθcosθ∫0lx2dx
τef=3ml2ω2sinθcosθ
τmg=τcf
mg.2lsinθ=3ml2ω2sinθcosθ
cosθ=2lω23g
JEE Main 2020 Physics, Rotational Motion — question figure A uniform rod of length ' ell ^' is pivoted at one of its ends on a vertical shaft of…
JEE Main 2020 — Physics Mechanics
2020mcqmedium
A uniform rod of length ' ℓ′ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θwith it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) 12mℓ2ω2sinθ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FHand Fv about the CM. The value of θ is then such that:
Official previous-year question
Held on 3 Sept 2020 · Verified 6 Jul 2026.
Options
- A
cosθ=3lω22g
- B
cosθ=2ℓω2g
- C
cosθ=ℓω2g
- D
cosθ=2ℓω23g
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