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Homework Help: Angular acceleration and Torque

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniform 6 kg rod with length 18 m has
    a frictionless pivot at one end. The rod is
    released from rest at an angle of 30◦ beneath
    the horizontal.What is the angular acceleration of the rod
    immediately after it is released? The moment
    of inertia of a rod about the center of mass
    is 1/12 mL2 , where m is the mass of the rod and L is the length of the rod. The moment
    of inertia of a rod about either end is 1/3 mL2, and the acceleration of gravity is 9.8 m/s2 .
    Answer in units of rad/s2.



    2. Relevant equations
    I(alpha), Torque


    3. The attempt at a solution
    I swear I've spent the past 3 hours searching for solutions online, my head, book, and nothing that has given me any answer that is reasonable. So, this is my last resort. :P

    From all of the types of ways I've found to solve it, this is the one I truly understand,

    since it is a rod with axis at the end it's inertia is 1/3m(length)^2,

    1/3m(length)^2*(alpha) = Torque= r*(Force)= (length)/2 *mg*cos(30)

    then solve for alpha,

    This is where I get confused... r*force is a cross product right? so shouldn't it be (Length)/2 * mg*sin(30)? and i don't even know if my right side of my equation is even right =.=\ does gravity even affect torque? it should... I've seen problems like this with a mass attached at the end to give extra torque and they don't even mention gravity acting upon the rod on the right side. I seriously won't sleep till I answer this!
     
  2. jcsd
  3. Apr 9, 2010 #2
    oh also, even made sure I was using the right moment of inertia by proving it with the parallel axis theorem
     
  4. Apr 9, 2010 #3
    Torque = I*alpha
    Torque = Force*Distance

    You know I, find torque, then solve for alpha.
    Draw a free body diagram and use trig to find the torque.

    The only force acting on the rod is gravity on its center of mass.

    The right side of your equation is correct. Draw a freebody diagram and you will see that
    the force of gravity normal to the bar in that position is mg*cos(30).
     
    Last edited: Apr 9, 2010
  5. Apr 9, 2010 #4
    I received 4.24352 rad/s^2.... for alpha.

    well, i already attempted my max amount of tries on it so I missed it :P but, I was able to verify it with another problem I have in my notes.

    Thanks again. you helped me notice that the cross product still holds F*r*sin(theta) but the theta is a different one, 90-30, which is totally obvious right? haha, i need to take a break =.=\
    Just what I get messed up on is which angle is which theta, the free body diagram helps alot, but even then I question myself :P. I need to remember from perpendicular to the rod to the vector of gravity is the same angle as the tilt from the horizontal. cheers! time to make a big glass of pepsi.
     
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