# Finding Angular Momentum

Tags:
1. Nov 30, 2015

### Mia

1. The problem statement, all variables and given/known data
A 2.0-m measuring stick of mass 0.175 kg is resting on a table. A mass of 0.500 kg is attached to the stick at a distance of 74.0 cm from the center. Both the stick and the table surface are frictionless. The stick rotates with an angular speed of 5.30 rad/s.

(a) If the stick is pivoted about an axis perpendicular to the table and passing through its center, what is the angular momentum of the system?

(b) If the stick is pivoted about an axis perpendicular to it and at the end that is furthest from the attached mass, and it rotates with the same angular speed as before, what is the angular momentum of the system?

2. Relevant equations
L=Iω
I=MR^2
I=(1/12)ML^2
I=(1/3)ML^2

3. The attempt at a solution

Okay so for the first question I used the moment of inertia formula and treated the mass as a point mass.
(0.500*0.740^2)=0.27
Plus the moment of inertia for the stick(rod) since it passing through the center I used I=(1/12)ML^2
(1/12)(0.175*2^2)=0.058
Then I added and multiplied by the angular velocity
0.27+0.058=0.33*5.3=1.7 kg*m/s

Now this answer was right however when I did the second one I repeated the same process but used I= (1/3)ML^2 because it at the end of the rod.

So I did
(0.500*0.740^2)=0.27
Then
(1/3)(0.175*2^2)=0.23
Then I added and multiplied by the angular velocity
0.27+0.23=0.50*5.3=2.65 rounded to 2.7 kg*m/s

It is saying both 2.65 and 2.7 were wrong. Where did I go wrong?

2. Nov 30, 2015

### (Ron)^2=-1

I think what you're missing here is the parallel axis theorem.

3. Nov 30, 2015

### Mia

Is that I=Icm +Md^2?

4. Nov 30, 2015

### (Ron)^2=-1

Yes it is.

5. Nov 30, 2015

### Mia

Okay so I would add that to the moment of inertia I had found for the stick then multiply by the angular velocity?

6. Nov 30, 2015

### (Ron)^2=-1

You need to find the total moment of inertia with respect to the new axis.

7. Nov 30, 2015

### Mia

Okay I'm getting a little confused.
So I=Icm +Md^2?
Would I use Icm = 1/12ML^2 for the center of mass?
thenMd^2
would d=1m since that would be the distance from the end of the rod to the center?

8. Nov 30, 2015

### (Ron)^2=-1

Yes, you're right! D is the distance from the centre of mass axis (axis of symmetry).

Always remember: The moment of inertia Icm is always with respect to the centre of mass, from which there is a symmetric distribution of mass (at least for homogeneous rigid bodies). So whenever you need to calculate the moment of inertia with respect to another parallel axis you can use the parallel axis theorem.

Last edited: Nov 30, 2015