A rod of mass 𝑚 and length L, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed 𝑣 strikes the rod horizontally at a distance 𝑥 from its pivoted end and gets embedded in it. The combined system now rotates with angular speed 𝜔 about the pivot. The maximum angular speed 𝜔𝑀 is achieved for 𝑥 $$=$$ 𝑥𝑀. Then

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Infinity Learn · Accepted Answer

by the angular momentum conservation about the suspension point.$$mvx=m$$ℓ$$23+mx2$$ω∴ω$$=mvxm$$ℓ$$23+mx2=3vx$$ℓ$$2+3x2$$ For maximum ω⇒dω$$dx=0$$⇒$$3vddxx$$ℓ$$2+3x2=0$$⇒$$ddx1$$ℓ$$2x+3x2x=0$$⇒ddxℓ$$2x+3x$$−$$1=0$$⇒$$($$−$$1)$$ℓ$$2x+3x$$−$$2$$−ℓ$$2x2+3=0$$⇒$$-$$ℓ$$2x2+3=0$$∴$$xM=$$ℓ$$3Thus$$, ω$$max=3v$$ℓ$$3$$ℓ$$2+3$$ℓ$$2/3=3V2$$ℓ So the ω$$=V2$$ℓ$$3$$

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