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jrrodri7
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1. A uniform horizontal rod of mass 3 kg and length 0.25 m is free to pivot about one end. The moment of inertia about an axis perpendicular to the rod and through the Center of Mass is given by I = [tex]\frac{ml^{2}}{12}[/tex]
If 8.9 force at an angle of 40 degrees to the horizontal acts on the rod, what's the magnitude of the resulting angular acceleration?http://img301.imageshack.us/img301/3267/torqueproblemscy6.jpg
2. Here's what I did.
[tex]\sum[/tex] [tex]\tau[/tex] = I [tex]\alpha[/tex] = ?
[tex]\tau[/tex] = LFsin[tex]\theta[/tex] = I [tex]\alpha[/tex]
Given an I, but the wrong I, I substituted the moment for a rod spinning on the edge, not the center of mass. Giving me...
[tex]\tau[/tex] = [tex]\frac{m^L{2}}{3}[/tex] [tex]\alpha[/tex]
I rearranged and plugged in my values for the alpha, and I ended up with...
1/3 [tex]\alpha[/tex] = [tex]\frac{Fsin\Theta}{ML}[/tex][tex]\alpha[/tex] = 3 * [tex]\frac{8.9 * sin (40)}{3 * 0.25}[/tex]
If 8.9 force at an angle of 40 degrees to the horizontal acts on the rod, what's the magnitude of the resulting angular acceleration?http://img301.imageshack.us/img301/3267/torqueproblemscy6.jpg
2. Here's what I did.
[tex]\sum[/tex] [tex]\tau[/tex] = I [tex]\alpha[/tex] = ?
[tex]\tau[/tex] = LFsin[tex]\theta[/tex] = I [tex]\alpha[/tex]
Given an I, but the wrong I, I substituted the moment for a rod spinning on the edge, not the center of mass. Giving me...
[tex]\tau[/tex] = [tex]\frac{m^L{2}}{3}[/tex] [tex]\alpha[/tex]
I rearranged and plugged in my values for the alpha, and I ended up with...
1/3 [tex]\alpha[/tex] = [tex]\frac{Fsin\Theta}{ML}[/tex][tex]\alpha[/tex] = 3 * [tex]\frac{8.9 * sin (40)}{3 * 0.25}[/tex]
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