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!