Mechanism of a cylinder rolling down a ramp?

  • Thread starter kendro
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Hi guys.... Can anyone hlp me with the mechanism of a cylinder rolling down a ramp? Here's my understanding about the problem. Can anybody corrected me?
If the CM is located at point O (the center of the cylinder), then the torque of for the rotational mechanism is provided by the friction force right? Then, the lever arm will be R (the radius of the cylinder). Then, the Torque will always be I (the moments of inertia) x alpha (angular acceleration)....

I got very confused when the CM is not located at the center of the cylinder, such as a distance d from point O. The weight of the CM will not in-line with the point where the cylinder is in contact with the ramp. Therefore, what is the normal force? Is it still W x cos(theta)? But then, if the CM is not located at the center of the cylinder, the weight of it will provide another torque other than the torque provided by the friction, with the lever arm of d x cos (beta)? But, then this torque will not be the same as the cylinder rolls down the ramp because the CM will change its position relative to the ramp, right? I am not sure about this....

Can anybody help to show me the correct force diagram for the attached diagram? Thank you very much guys..... I really appreciate any kind of help....
 

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Answers and Replies

  • #2
quasar987
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Here's my humble take on it.

There's a torque provided by gravity, another provided by friction and another by normal force. All with lever d because of the theorem "when many forces act on a solid body of mass M, the motion of the CM is that of a particle of mass M on which all the forces on the body are exerted"
 

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