Difficult energy conservation/rotational energy problem

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SUMMARY

The discussion centers on a physics problem involving energy conservation and rotational dynamics, specifically a system with a hanging mass (m1 = 0.405 kg), a sliding block (m2 = 0.825 kg), and a hollow cylindrical pulley (M = 0.350 kg) with inner radius R1 = 0.020 m and outer radius R2 = 0.030 m. The coefficient of kinetic friction is μk = 0.250, and the block has an initial velocity of vi = 0.820 m/s. The main challenge identified is calculating the moment of inertia for the hollow pulley, which requires understanding the combination of both radii and potentially applying the parallel axis theorem.

PREREQUISITES
  • Understanding of kinetic and potential energy equations (KE and PE)
  • Knowledge of moment of inertia for hollow cylinders
  • Familiarity with the parallel axis theorem
  • Basic principles of friction in mechanical systems
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  • Calculate the moment of inertia for a hollow cylinder using both inner and outer radii
  • Review the application of the parallel axis theorem in rotational dynamics
  • Explore the relationship between kinetic energy, potential energy, and work done by friction
  • Practice solving similar energy conservation problems in mechanics
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of rotational dynamics problems.

doneky
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Homework Statement


In the figure below, the hanging object has a mass of m1 = 0.405 kg; the sliding block has a mass of m2 = 0.825 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface isμk = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table.
10-p-049.gif


Homework Equations


KE = 1/2Iw^2
KE = 1/2mv^2
PE = mgh
Non-conservative work = (delta)KE + (delta)PE

The Attempt at a Solution


The left side has negative work because of friction. On the right side, I put the KE of each as well as the potential energy (ONLY on m1, I believe?)

I think my only problem is the moment of inertia for the two pulleys. I don't understand how to implement the two radii. The answer shows that there is a combination of both. I have no idea why. Parallel axis theorem maybe?
 
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doneky said:
I think my only problem is the moment of inertia for the two pulleys.
There is only one pulley. It has a hollow cylinder with inner radius ##R_1## and outer radius ##R_2##. You need to compute its moment of inertia ##I##.

You also have not stated the actual problem, i.e., what you are asked to find, only the setup.
 

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