Clarification regarding rolling cylinder with CG slightly above

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The discussion revolves around the work done by rolling friction and the normal reaction for a solid cylinder with its center of gravity slightly above its geometric center while rolling down an inclined plane. The user seeks clarification on the calculations for the work done by rolling friction, proposing W1 = F * c * Sinθ, and the work done by the normal reaction, suggesting W2 = R * (c - c * Cosθ). They express uncertainty about the correctness of these equations, particularly noting that for a cylinder with its center at O, the work done by the normal reaction is zero. The thread invites input from others to confirm or correct these calculations. Overall, the focus is on understanding the dynamics of the rolling cylinder in relation to its center of gravity.
seshikanth
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Hi

Can anyone please clarify the following?

Let there by a rolling cylinder on an inclined plane. Let the solid cylinder(with Center O) on the inclined plane has Center of gravity at a distance 'c' slightly above O. If the cylinder is displaced slightly through an angle θ then

[Let N = normal reaction; F= Rolling friction]

1. What is the work done by rolling friction F
I think, W1 = F. c Sinθ. ---- Can you tell am i correct here?

2. What is the work done by Normal Reaction?
I think it is W2 = R. (c - c Cosθ) ; --- Can you tell me if i am correct here? I know that the work done by normal reaction for a cylinder with center O as zero !



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