Kinetic energy around an axis - two methods?

[Niles]

Homework Statement

I have to find the kinetic energy of the total system; b has mass 2m and a has mass m. They rotate around the y-axis and and through the origin (0,0).

There's two ways of finding the kinetic energy of the system around the y-axis.

1) I find moment of inertia for each particle around the y-axis (and through the origin) and add - so it's 2*m*r^2+9*m*r^2 = 11*m*r^2. Then I use this in K = ½*I*w^2.

2) I find the center of mass: r_cm = (2*m*r+m*3*r)/(3m) = (5/3)r. I use this in I = m*r^2 and then use K = ½*I*w^2.

#2 doesn't work - why?

 

[Niles]

Wait a minute.. when I use #2, I find the moment of inertia I_cm, not I_y, right?

 

[Niles]

I have another question.

If I want to find the kinetic energy, I can do it like this:

K = ½*I_total*w^2, where w^2 = v^2/r^2 - what radius are we talking about here? Is it the distance from center of mass to the point on the axis, on which they are turning?

Or can I only use K = ½*I_single*v^2/r^2 when looking at the particles individually, and then add the two energies? Here r is the perpendicular distance to the axis.

 

[Doc Al]

Homework Statement

I have to find the kinetic energy of the total system; b has mass 2m and a has mass m. They rotate around the y-axis and and through the origin (0,0).

There's two ways of finding the kinetic energy of the system around the y-axis.

1) I find moment of inertia for each particle around the y-axis (and through the origin) and add - so it's 2*m*r^2+9*m*r^2 = 11*m*r^2. Then I use this in K = ½*I*w^2.

2) I find the center of mass: r_cm = (2*m*r+m*3*r)/(3m) = (5/3)r. I use this in I = m*r^2 and then use K = ½*I*w^2.

#2 doesn't work - why?

When analyzing rotation of an extended body, you can't just replace the body by a point mass at its center of mass. (Imagine a body rotating about an axis through its center of mass. It has some rotational inertia. But replace it by a point mass at that point and it has none.)

I have another question.

If I want to find the kinetic energy, I can do it like this:

K = ½*I_total*w^2, where w^2 = v^2/r^2 - what radius are we talking about here? Is it the distance from center of mass to the point on the axis, on which they are turning?

Or can I only use K = ½*I_single*v^2/r^2 when looking at the particles individually, and then add the two energies? Here r is the perpendicular distance to the axis.

r is always the perpendicular distance to the axis. In the first case, for some reason you aren't given w but have the speed of some point, so you use that point's distance from the axis to calculate w.

 

[Niles]

Ok, so I can't find the kinetic energy of the total system by using

K = ½ * I_{total} * v^2/r^2? If I use this equation, I have to find it for each particle, and then add it?

My problem here is: In "K = ½ * I_{total} * v^2/r^2", what is r for the total system?

 

[Doc Al]

What you want to use is:
$$K = 1/2 I_{total} \omega^2$$

How you find $\omega$ depends on what you're given.

If you happen to have the speed of one particular point of the rotating system, you can use it to find omega. r would not be for "the total system", but just for that particular point.

 

[Niles]

Ok, I still don't believe my question has been answered.

If I want to calculate the kinetic energy of the total system, I do like this:

K_total = ½*I_a*w_a + ½*I_b*w_b

The total moment of inertia is 11*m*r^2, and we call this I_total. The total kinetic energy, K_total, can also be found as

K_total = ½*I_total*w_system.

In w_system, what is the distance r? That is my question.

 

[Niles]

Ok, when reading your replies, my question has been answered.

Thank you very much!