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) $$ml212$$ω$$2sin$$θ$$cos$$θ 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 :

Question

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)  $$ml212$$ω$$2sin$$θ$$cos$$θ 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 :

Infinity Learn · Accepted Answer

From a rotating frame rod will appear in equilibrium. Net torque about suspension point must be zero.τC.$$F=$$τmgNow, torque due to $$gravity=$$τ$$mg=mg$$ℓ$$2sin$$θtorque due to centrifugal $$force=$$τC.$$F=$$∫$$FCF(xsin$$θ$$)=$$∫$$dm(xcos$$θ$$)$$ω$$2xsin$$θ$$=$$∫mℓdx $$x2$$ ω$$2cos$$θ$$sin$$θ $$=m$$ℓ$$23$$ω$$2cos$$θ$$sin$$θ ∴mgℓ$$2sin$$θ$$=m$$ℓ$$2$$ω$$2cos$$θ$$sin$$θ$$3$$ ⇒ $$cos$$θ$$=3$$ $$g2$$ℓω$$2$$

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