# Rotation of a rigid body

1. Nov 19, 2012

### Toranc3

1. The problem statement, all variables and given/known data

A 0.9 kg mass at (x, y) = (20 cm,20 cm) and a 2.0 kg mass at (20 cm,100 cm) are connected by a massless, rigid rod. They rotate about the center of mass.

2. Relevant equations

x=(m1*x1+m2*x2)/(m1+m2)

y=(m1*y1+m2*y2)/(m1+m2)

I=1/12*ML^(2)
3. The attempt at a solution

What are the coordinates of the center of mass?
I got 0.20m and 0.752m

What is the moment of inertia about the center of mass?

I used this 1/12*(m1+m2)*(L1x+L2x+L1y*L2y)^(2)
1/12*(0.9kg+2.0kg)*(.552-.248)^(2)

I get 0.02233kg*m^(2) but this is wrong. The answer is 0.397kg*m^(2)

What am I doing wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 19, 2012

### TSny

This is the formula for the moment of inertia of a rod with mass. The rod in your problem is massless.

You have two point particles. So, you need to know how to determine the moment of inertia of a point particle.

3. Nov 19, 2012

### Toranc3

Ah I see. Since there is a y component how would I go about doing that? would I have to subtract each of the masses components with the components of the center of mass?

4. Nov 19, 2012

### TSny

If you're talking about finding the distance of each mass from the axis of rotation (center of mass point), then yes.

5. Nov 19, 2012

### Toranc3

Thanks! :)

6. Nov 19, 2012

### Toranc3

What if the question gave a mass for the rod? I would then use the rods interia formula and add the rods inertia to the masses inertia? would that be correct?

7. Nov 19, 2012

### TSny

Yes. But there's a complication in that the equation $I = (1/12) M L^2$ assumes that you rotate about the center of mass of the rod. If you rotate about some other point (such as the center of mass of the whole system) then you would need to use a different formula for $I$ of the rod.

8. Nov 19, 2012

### Toranc3

I have another question sorry. At what angle with respect to that axis of the rod should 1.2N forces be applied to each mass to give the torque you find in part c?

Torque= 0.828N*m

I am assuming this formula will be used

Torque=Frsin(theta) im stuck with my r.

9. Nov 19, 2012

### TSny

By definition, r is the distance from the axis of rotation to the point of application of the force.

10. Nov 19, 2012

### Toranc3

Man every time you answer my question I keep going "oooohh" I should have known that. I am not thinking straight today sorry. But thank you so much for your help. :).

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