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USING ENERGY CONSERVATION ON THE PARTICLE ONLY WE CAN THE SPEED WHEN IT IS AT THE BOTTOM POSITION.. BUT THE SOLUTION WHICH I SAW WAS A LENGTHY ONE CHANGE IN PE=2*5/4R=CHANGE IN KINETIC ENERGY=1/2*MV^2 HENCE V=(5GR)^1/2 PLZ CONFIRM.. 4. A uniform circular disc has radius R and mass m. A particle, also of mass m, is fixed t a point A on the edge of the disc as shown in the 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 the point A at its highest position. It is the then allowed to fall so that it starts rotating about PQ. Find the linear speed of the particle as it reaches its lowest position.
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