Solve for Angular Acceleration: Torque Problem on a Uniform Rod

[jrrodri7]

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 = \frac{ml^{2}}{12}

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.

\sum \tau = I \alpha = ?

\tau = LFsin\theta = I \alpha

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...

\tau = \frac{m^L{2}}{3} \alpha

I rearranged and plugged in my values for the alpha, and I ended up with...

1/3 \alpha = \frac{Fsin\Theta}{ML}\alpha = 3 * \frac{8.9 * sin (40)}{3 * 0.25}

 

[tron_2.0]

if you break the force vector into components, you get that the horizontal compontant is Fcos*theta and the vertical component is Fsin*theta

in your torque diagram, the weight of the rod is at the center (r=l/2) and causes the rod to want to go clockwise. the vertical component of the force vector (Fsin*theta) wants to make the rod go counterclockwise.

this means that the sum of the torques should equal: (Fsin*theta*l)-(m*g*l/2)

its l/2 because the weight of the rod comes from the center of the rod, at a radius of l/2

 

[jrrodri7]

So, given that I can apply the same idea I was going for to find out Alpha and set the sum of the torques equal: I * alpha (angular acceleration), or would I have to do two separate equations and find a resultant vector for the acceleration plugging in the two separate components of "F" into that equation you have presented.

 

[jrrodri7]

So Just set the radii to (0.25/2) because at w_r you are at the CoM? This would give something along the lines of T = w_r - Fsin(40) and set that to I alpha? replacing the radii in I with (0.25/2)?