Find an Expression for the Frequency - Pendulum

Click For Summary

Homework Help Overview

The problem involves finding an expression for the frequency of small angle oscillations of a solid sphere suspended from a thin rod. The discussion centers around the interpretation of the variables in the frequency equation and the moment of inertia relevant to the setup.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions of R and I in the frequency equation, questioning whether R refers to the radius of the sphere or the distance from the pivot to the center of mass. There is also discussion about the appropriate moment of inertia to use in the context of the problem.

Discussion Status

Participants are actively questioning the assumptions made regarding the variables in the frequency equation. Some have provided clarifications about the definitions of R and I, while others are seeking further visual context to understand the setup better. No consensus has been reached yet.

Contextual Notes

There is ambiguity in the problem regarding the arrangement of the rod and the sphere, which affects the interpretation of R and I. Participants are considering different interpretations of the setup and the implications for the calculations involved.

AbigailG
Messages
12
Reaction score
1

Homework Statement


[/B]
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


[/B]
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?
 
Physics news on Phys.org
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.
 
TSny said:
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?
 
Is the setup like either of the pictures below?

upload_2017-12-11_14-49-22.png
 

Attachments

  • upload_2017-12-11_14-49-22.png
    upload_2017-12-11_14-49-22.png
    4.4 KB · Views: 2,200
AbigailG said:
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
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
17
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K