Analyzing Simple Harmonic Motion of a Rolling Sphere in a Cylindrical Trough

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The discussion focuses on analyzing the motion of a solid sphere rolling without slipping in a cylindrical trough, demonstrating that it exhibits simple harmonic motion (SHM) for small displacements. The period of this motion is derived as T = 2Pi √(28R/5g). The approach involves treating the sphere's motion similarly to that of a physical pendulum, with considerations for the center of mass and the moment of inertia. The challenge lies in correctly calculating the moment of inertia and understanding the relationship between angular speed and angular frequency. The conversation emphasizes the need for clarity in the dynamics of rolling motion to accurately apply SHM principles.
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Homework Statement


A solid sphere (radius = R) rolls without slipping in a
cylindrical trough (radius = 5R), as shown in Figure
P13.56. Show that, for small displacements from equilib-
rium perpendicular to the length of the trough, the
sphere executes simple harmonic motion with a period
T = 2Pi √28R/5g.



Homework Equations





The Attempt at a Solution


It's essentially a pendulum type problem except a ball is rolling instead of just moving. Once we get w were set:
Using physical pendulum, w = root(mgd/I):
d = distance to CM of system = 5R-R = 4R = CM of ball.
I = I was thinking that the axis of rotation is at 4R from the CM of the ball, Icm of ball = 2/5MR^2, so I was thinking I = 2/5MR^2 + M(4R)^2, but this doesn't look like it's going to put me in the right direction.

Any thoughts?
 
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The ball rotates around its own CM while rolling in the trough and the CM performs circular motion around the axis of the cylinder. The angular speed of rotation is different from the angular frequency of the SHM.

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