Physics Help Forum equation of motion

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 Jul 12th 2017, 06:10 PM #1 Junior Member   Join Date: Feb 2017 Posts: 5 equation of motion Could someone explain equation 2? (the part with the red frame) Thank you Attached Thumbnails
 Jul 13th 2017, 05:40 AM #2 Senior Member     Join Date: Aug 2008 Posts: 103 $\tau_{net} = J_0 \alpha = Tr + F_s R$ 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 07:36 AM.
 Jul 13th 2017, 07:22 AM #3 Junior Member   Join Date: Feb 2017 Posts: 5 Thank you very much. Does the second 4r refer to θ?
Jul 13th 2017, 07:35 AM   #4
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Join Date: Aug 2008
Posts: 103
 Originally Posted by cxz 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 07:37 AM.

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