A uniform circular disc has radius R and mass m. A particle, also of mass m, is fixed at point A on the edge of the disc as shown in figure. The disc can rotate freely about a fixed horizontal chord PQ that is at a distance $$R/4$$ from the centre C of the disc. The line AC is perpendicular to PQ. Initially, the disc is held vertical with point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. If R $$=$$ $$0.5$$ m then, find the linear speed of the particle (in$$ $$m/s) as it reaches its lowest position. (Take$$ g $$=$$ $$10$$ $$m/s2)

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Infinity Learn · Accepted Answer

The system (disc$$ $$+$$ $$particle) is rotating about fixed axis without friction, its mechanical energy should be conserved. According to conservation of energy, Δ$$K+$$Δ$$U=0$$⇒Kf−$$Ki+$$Δ$$U=0$$        …(i)During$$ the fall, the $$disc-mass$$ system gains rotational kinetic energy. This is at the expense of potential energy.Change in gravitational potential energy.Δ$$U=$$−$$mg2R+R4+mg2R4=$$−$$3mgRChange$$ in kinetic energy Δ$$K=12IPQ$$ω$$2$$−$$0where$$ $$IPQ=MI$$ of the $$disc-mass$$ system about $$PQIPQ=Idisc$$ $$PQ+Imass$$ $$PQ(I)PQ=Idisc$$ $$PQ+Imass$$ $$PQ=mR24+MR42+m5R42=15mR28$$Δ$$K=1215mR28$$ω$$2$$−$$0Finally$$ substituting ΔU and ΔK in the equation $$(i),we $$have1215mR28$$ω$$2$$−$$0+($$−$$3mgR)=0$$⇒ω$$=16g5RLet$$ v be the velocity of mass m at the lowest point of rotation $$v=$$ω$$R+R4=16g5R$$×$$5R4=5gR=5$$×$$10$$×$$0.5=5m/s$$

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