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

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In summary, a solid sphere is rolling without slipping in a cylindrical trough and exhibits simple harmonic motion for small displacements perpendicular to the length of the trough. The period of the motion is determined to be 2Pi √28R/5g. The attempt at a solution involves using the physical pendulum equation and calculating the distance to the center of mass of the system and the moment of inertia. However, there is a distinction between the angular speed of rotation and the angular frequency of the SHM.
  • #1
AriAstronomer
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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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  • #2
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.

ehild
 

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

1) How do you define "tough oscillation"?

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