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.
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!