Calculating Angular Acceleration for a Pully Problem with a Solid Cylinder

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Homework Help Overview

The discussion revolves around calculating the angular acceleration of a solid cylinder in two scenarios: one where a force equal to the weight of a mass is applied, and another where the mass itself is hung from the string. The subject area involves concepts of rotational dynamics and forces.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the difference between applying a force versus hanging a mass, questioning how the acceleration of the mass affects the system. There is a suggestion to consider the relationship between linear and angular quantities.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the implications of the two scenarios. Some guidance has been offered regarding the dynamics of the system, but there is no explicit consensus on the next steps or solutions.

Contextual Notes

Participants note that the mass provides a constant force only if it is not accelerating, which raises questions about the assumptions in the problem setup. There is also mention of a differential equation relating to the system's dynamics.

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M, a solid cylinder (M=1.39 kg, R=0.137 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.710 kg mass, i.e., F = 6.965 N. Calculate the angular acceleration of the cylinder.

That I can do... But then I get...

If instead of the force F an actual mass m = 0.710 kg is hung from the string, find the angular acceleration of the cylinder.

Why would the second situation be different than the first? And could you point me in the right direction? Thanks.
 
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ninjagowoowoo said:
...A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.710 kg mass If instead of the force F an actual mass m = 0.710 kg is hung from the string, find the angular acceleration of the cylinder.
The 0.71 kg mass provides a constant downward force only if it's not accelerating, itself.
 
I get what you're saying, but I still have no idea where to go from here. Do you think you could give me some more hints? Thanks for the help. :confused:
 
I have to leave this computer pretty soon, but here's a quick suggestion: The D.E. for the system will have a mass with a linear acceleration and a force, and a moment of inertia with an angular acceleration and a torque. The torque & ang. accel. can be converted to linear units of force & linear accel, and will equal the tension in the line and the acceleration of the mass. Gotta go...
 

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