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Rotational SHM

  • Thread starter suspenc3
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https://www.physicsforums.com/showthread.php?p=846415

Can anyone explain this a little bit more?

I found these 2 formulas.
[tex]\tau=-k \theta[/tex]

How do I relate [tex]\theta[/tex] with m, R, r & k? (In my problem im not working with numbers, i need to find an expression for angular frequency in terms of R, r, m, k,

[tex]\tau=I \alpha[/tex] where [tex]I=MR^2[/tex]

therefore [tex]\alpha = \frac{-k_t\theta}{MR^2}[/tex]

Do I just resolve [tex]\theta[/tex] into its horizontal component? That wouldnt really get rid of [tex]\theta[/tex] though...Confused
Thanks
 
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  • #2
OlderDan
Science Advisor
Homework Helper
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https://www.physicsforums.com/showthread.php?p=846415

Can anyone explain this a little bit more?

I found these 2 formulas.
[tex]\tau=-k \theta[/tex]

How do I relate [tex]\theta[/tex] with m, R, r & k? (In my problem im not working with numbers, i need to find an expression for angular frequency in terms of R, r, m, k,

[tex]\tau=I \alpha[/tex] where [tex]I=MR^2[/tex]

therefore [tex]\alpha = \frac{-k_t\theta}{MR^2}[/tex]

Do I just resolve [tex]\theta[/tex] into its horizontal component? That wouldnt really get rid of [tex]\theta[/tex] though...Confused
Thanks
From the diagram in the link you posted you should be able to write an expression for the elongation (or compression) of the spring Δx in terms of the small angular displacement of the wheel from the equilibrium position. The force applied at the point of connection between the spring and the wheel is proportional to Δx. The torque about the axle of the wheel is the result of that force applied at distance r from the axis of rotation.

The "k" in your equation above is not necessarily the spring constant of the spring in the diagram. You need to find the proportionality constant between torque and angular displacement.
 

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