MI moment of inertia on pulley

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SUMMARY

The discussion centers on calculating the acceleration of two masses, m1 and m2, connected by a cord over a pulley with a moment of inertia of 5 kg·m² and a radius of 0.5 m. The initial approach incorrectly assumed total torque as 7g(0.5), leading to an erroneous angular acceleration calculation. The correct method requires considering the opposing forces exerted by the two masses, which affects the net torque acting on the pulley. The realization of this mistake highlights the importance of accurately accounting for opposing forces in rotational dynamics.

PREREQUISITES
  • Understanding of Newton's second law for rotation
  • Familiarity with the concepts of torque and moment of inertia
  • Basic knowledge of angular acceleration and linear acceleration relationships
  • Experience with frictionless pulley systems
NEXT STEPS
  • Study the principles of torque in rotational dynamics
  • Learn how to apply Newton's second law to systems involving pulleys
  • Explore the effects of opposing forces on net torque calculations
  • Investigate the relationship between linear and angular motion in mechanical systems
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Physics students, mechanical engineers, and anyone interested in understanding the dynamics of pulley systems and rotational motion.

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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 a lot! How could I have made this mistake?! ARGH!
 

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