Find an Expression for the Frequency - Pendulum

[AbigailG]

Homework Statement

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A solid sphere of mass M and radius R is suspended from a thin rod. The sphere can swing back and forth at the bottom of the rod. Find an expression for the frequency of small angle oscillations.

Homework Equations

f = 1/2(pi) sqrt(MgR/I)

I for a solid sphere 2/5MR^2

The Attempt at a Solution

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I simply plugged the moment of inertia into the equation for the frequency and that yielded:

1/2(pi) sqrt(5g/2R)

but the answer is:

1/2(pi) sqrt(5g/7R)

Does anyone know what my missing is?

 

[TSny]

Review the meaning of R and I in the equation f = 1/2(pi) sqrt(MgR/I). R is not the radius of the sphere and I is not the moment of inertia of the sphere about an axis through the center of the sphere.

EDIT: It could be that R does in fact equal the radius of the sphere depending on the interpretation of the problem. It's not clear how the rod is arranged in the problem.

 

[AbigailG]

Review the meaning of R and I in the equation f = 1/2(pi) sqrt(MgR/I). R is not the radius of the sphere and I is not the moment of inertia of the sphere about an axis through the center of the sphere.

Generically, R = l which is the distance from the pivot point to the center of mass. I thought the moment of inertia would be that of a sphere about an axis through the center. ..am I just not visualizing this correctly?

 

[TSny]

Is the setup like either of the pictures below?

 

[TSny]

Generically, R = l which is the distance from the pivot point to the center of mass.

Yes

I thought the moment of inertia would be that of a sphere about an axis through the center.

No, check out this page and pay particular attention to the meaning of $I_{\rm support}$.

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html