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Homework Help: Torque problem (rod)

  1. Mar 21, 2008 #1
    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 [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]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 21, 2008 #2
    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
  4. Mar 22, 2008 #3
    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.
  5. Mar 22, 2008 #4
    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)???
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