Angular Acceleration of Poll: Find the Equation

[asi123]

Homework Statement

Hey guys.
I got this poll which is at one side connected to a wall and is being released from a horizontal state.
his length is L and mass m.
I need to find the angular acceleration of the poll as a function of the angle.
I got this answer in the pic but according to them the answer should be what I marked in the red box.

Any idea? am I wrong?

10x.

Homework Equations

The Attempt at a Solution

 

[Doc Al]

You are using the parallel axis theorem to find the rotational inertia, but 1/3mL² is the rotational inertia about the axis (what you need) not the center of mass.

 

[asi123]

You are using the parallel axis theorem to find the rotational inertia, but 1/3mL² is the rotational inertia about the axis (what you need) not the center of mass.

Oh, stupid me, 10x.

 

[asi123]

Another thing.
First of all, when they say this is the angular acceleration of the poll, does it mean the angular acceleration of the center of mass?
In the next part of the question, they ask for the angular velocity as a function of \varphi.
I tried to use energy conservation and came up with this equation:
mgL=mgL(1-1/2\cos(\varphi))+1/2 mv^2+1/2 I\omega^2
When
v = \omega*(L/2)

Is this right?

10x.

 

[Doc Al]

First of all, when they say this is the angular acceleration of the poll, does it mean the angular acceleration of the center of mass?

All parts of the pole have the same angular acceleration.

In the next part of the question, they ask for the angular velocity as a function of \varphi.
I tried to use energy conservation and came up with this equation:
mgL=mgL(1-1/2\cos(\varphi))+1/2 mv^2+1/2 I\omega^2
When
v = \omega*(L/2)

Is this right?

Not exactly:

1. Where are you measuring PE with respect to? (Where does PE = 0?) What's the initial PE?
2. Since the angle is measured from the horizontal, revisit how the PE varies with angle.
3. Be careful not to count KE twice. The pole can be considered to be in pure rotation about the pivot, so 1/2 I\omega^2 represents the entire KE.

 

[asi123]

All parts of the pole have the same angular acceleration.

Not exactly:

1. Where are you measuring PE with respect to? (Where does PE = 0?) What's the initial PE?
2. Since the angle is measured from the horizontal, revisit how the PE varies with angle.
3. Be careful not to count KE twice. The pole can be considered to be in pure rotation about the pivot, so 1/2 I\omega^2 represents the entire KE.

Is this right?

 

[Doc Al]

Is this right?

Yes. That looks good.

 

[asi123]

10x.