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Old Jul 12th 2017, 05:10 PM   #1
cxz
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equation of motion

Could someone explain equation 2? (the part with the red frame)
Thank you
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Old Jul 13th 2017, 04:40 AM   #2
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$\tau_{net} = J_0 \alpha = Tr + F_s R$

Hooke's law for an ideal linear spring says ...

$F_s = -kx$

recall arc length equals radius times angular displacement ...

$x = R \cdot \Delta \theta$

$R = 4r$ and $\Delta \theta = \theta_0 + \theta$, where $\theta_0$ is the angular displacement of the pulley at system equilibrium, and $\theta$ is the additional angular displacement beyond equilibrium to induce an oscillation of the system.

$F_s R = -k \color{blue}{x} \cdot \color{red}{R} = -k \cdot \color{blue}{4r(\theta_0+\theta)} \cdot \color{red}{4r}$

Last edited by skeeter; Jul 13th 2017 at 06:36 AM.
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Old Jul 13th 2017, 06:22 AM   #3
cxz
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Thank you very much.

Does the second 4r refer to θ?
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Old Jul 13th 2017, 06:35 AM   #4
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Originally Posted by cxz View Post
Thank you very much.

Does the second 4r refer to θ?
The second $4r$ is the radius in the torque formula. The first $4r$ is used to calculate the arc length, $x$.

Check my last post again ... I color-coded it.

Last edited by skeeter; Jul 13th 2017 at 06:37 AM.
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