Finding moment of inertia for two masses on a rod rotating about their center of mass

[Toranc3]

Homework Statement

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.

Homework Equations

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

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

I=1/12*ML^(2)

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?

 

[TSny]

I=1/12*ML^(2)

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.

 

[Toranc3]

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.

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?

 

[TSny]

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?

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

 

[Toranc3]

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

Thanks! :)

 

[Toranc3]

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

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?

 

[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.

 

[Toranc3]

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

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) I am stuck with my r.

 

[TSny]

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

 

[Toranc3]

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

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. :).