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1. A uniform horizontal rod of mass

If

http://img301.imageshack.us/img301/3267/torqueproblemscy6.jpg [Broken]

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]

**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 [Broken]

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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