# Angular acceleration and Torque

## Homework Statement

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 .

I(alpha), Torque

## 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!

oh also, even made sure I was using the right moment of inertia by proving it with the parallel axis theorem

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).

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