A pulley has a moment of inertia of 5kg metre squared and a radius of 0.5m. The cord suporting the masses m1 and m2 does not slip and is hung on each side of the pulley with a cord thru the pulley. Assume that the axle is frictionless. How to find the acceleration of each mass when m1=2 and m2=5?(adsbygoogle = window.adsbygoogle || []).push({});

Why cant the method be as follows Total torque=Moment of inertia * angular accel

hence, 7g(0.5)=5alpha and alpha =0.7g. since a=r(alpha) then we know r=0.5 so we get a=3.43 but why is this not the correct ans?

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# Homework Help: MI moment of inertia on pulley

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