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 θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) ml212ω2sinθcosθ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of θ is then such that :
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a
cosθ=3g2lω2
b
cosθ=2g3lω2
c
cosθ=g2lω2
d
cosθ=glω2
answer is A.
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Detailed Solution
From a rotating frame rod will appear in equilibrium. Net torque about suspension point must be zero.τC.F=τmgNow, torque due to gravity=τmg=mgℓ2sinθtorque due to centrifugal force=τC.F=∫FCF(xsinθ)=∫dm(xcosθ)ω2xsinθ=∫mℓdx x2 ω2cosθsinθ =mℓ23ω2cosθsinθ ∴mgℓ2sinθ=mℓ2ω2cosθsinθ3 ⇒ cosθ=3 g2ℓω2
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 θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) ml212ω2sinθcosθ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of θ is then such that :