Pulley system with Moment of Inertia

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SUMMARY

The discussion centers on a pulley system involving two blocks with masses m1 = 2.05 kg and m2 = 5.95 kg, connected by a massless string over a solid disk pulley with radius R = 0.250 m and mass M = 10.0 kg. The blocks operate on a fixed block-wedge inclined at 30 degrees, with a coefficient of kinetic friction of 0.360. Key equations provided include the forces acting on both blocks and the moment of inertia for the pulley, expressed as I = MR², where F_tension * R = I * alpha.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with the concept of moment of inertia
  • Knowledge of free-body diagrams
  • Basic principles of friction in physics
NEXT STEPS
  • Calculate the moment of inertia for the pulley using I = MR²
  • Analyze the forces acting on both blocks using free-body diagrams
  • Determine the acceleration of the system using the provided equations
  • Explore the effects of varying the coefficient of kinetic friction on system dynamics
USEFUL FOR

Physics students, mechanical engineers, and anyone studying dynamics and rotational motion in pulley systems will benefit from this discussion.

TrippingBilly
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A block of mass m1 = 2.05 kg and a block of mass m2 = 5.95 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30.0°. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley. (The mass m2 is sitting on a 30 degree slope with the horizontal, which is connected by a string to the pulley at the top of the slope, and m1 is on the other side of the pulley on a flat surface.)

Before I can start to work on this problem, I'm not sure what the equation for the moment of inertia of the pulley would be. I don't know what direction to head in once I figure that but I can give it a try before I come back for more help.
 
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M1: (m1)(g)sin(30°) + (m1)(a) - (m1)(μk)(g)cos(30°) = 0M2: (m2)(g)sin(30°) + (m2)(a) - (m2)(μk)(g)cos(30°) = 0Pulley:I = MR^2F_tension * R = I * alpha
 

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