MI moment of inertia on pulley

AI Thread Summary
To find the acceleration of the masses m1 and m2 on a pulley with a moment of inertia of 5 kg·m² and a radius of 0.5 m, the total torque must account for the opposing forces of the two masses. The initial calculation incorrectly assumed a single torque value, leading to an incorrect angular acceleration. Instead, the correct approach involves calculating the net torque from both masses, which pull in opposite directions. This realization clarifies the mistake in the initial torque equation. Understanding the dynamics of the system is crucial for accurate calculations.
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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?

Why can't 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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Realize that the total torque is not 7g(0.5): the two masses pull in opposite directions.
 
Oh thanks alot! How could I have made this mistake?! ARGH!
 

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