Rotational Motion- Pulleys with Friction

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SUMMARY

The discussion centers on calculating the acceleration of two masses suspended from a pulley with friction. The pulley has a mass of 0.20 kg and a radius of 0.15 m, generating a constant torque of 0.35 MN due to friction. The masses are m1 = 0.4 kg and m2 = 0.8 kg. The derived formula for acceleration is a = (m2 - m1)g / (1/2M + m1 + m2), where the challenge lies in incorporating the effects of torque on the system.

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  • Understanding of Newton's Second Law of Motion
  • Familiarity with torque and moment of inertia concepts
  • Knowledge of rotational dynamics and angular acceleration
  • Basic algebra for solving equations involving multiple variables
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Homework Statement


Two Masses are suspended from a pulley. The pulley itself has a mas of .20kg a radius of .15m a constant torque of .35 MN due to the friction between the rotating pulley and its axle. What is the magnitude of the acceleration of the suspended masses if m1= .4kg and m2= .8 kg (neglect the mass of the string)
Pulley
| |
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M2 M1




Homework Equations


T1= M1G - M1A
T2= M2G + M2A
Torque= FL = Moment of Inertia x Alpha,
I= 1/2 mr^2

The Attempt at a Solution


I have tried solving this by solving for A when T2-T1= I alpha, but i really don't know how to apply the fact that the tension is different of both cords due to the friction on the pulley I am stuck...
 

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I have tried everything... but still i can't get to the answer.. i get to the formula a=(m2-m1)g/(1/2M+m1+m2) but after that i don't know how to apply the torque... :( need help
 
Last edited:
bump...
 

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