# Rotational dynamics

1. Jul 16, 2016

### Clara Chung

1. The problem statement, all variables and given/known data
A uniform bar AB of length L is freely hinged at one end A and released from a horizontal position.
a) Find the initial angular momentum. Ans: 3g/2L
b)Find the angular velocity when the bar is vertical. Ans: (3g/L)^(1/2)

2. Relevant equations
momentum of inertial of a rod through end =(ML^2)/3
3. The attempt at a solution
I have no problem with part a, I did it like this
Let N be the moment, I be the inertia of moment, a be the angular acceleration.
N = Ia
mg(L/2) = ML^2 /3 a
a= 3g/2L

But I have problem with part b, I tried to do it with energy conservation
Kr + Kt = PE
Iw^2 /2 + mw^2(L^2)/2 = mgL/2
1/3 L w^2 + w^2L = g

I would be correct if I neglect the linear KE, but I don't think it should be neglected, is my method of doing part b wrong?

2. Jul 16, 2016

### TSny

For rotation about a fixed axis, A, the total KE of the object is (1/2)IAω2.

It is possible to express this as KE = (1/2)Mvcm2 + (1/2)Icmω2. But this would not be helpful for this problem.

3. Jul 16, 2016

### Clara Chung

Thanks. what is Icm?

4. Jul 16, 2016

### TSny

cm is center of mass

5. Jul 16, 2016

### TSny

Moment of inertia, I, depends on the origin. The moment of inertia about the center of the rod is different than the moment of inertia about one end of the rod.