# Kinetic energy and where it rotates

1. Apr 1, 2004

### ShawnD

I got a test back and the teacher marked my entire answer wrong because to get the kinetic energy, I used the moment of inertia which included the parallel axis theorum to account for where it's rotating. The teacher says rotational energy is not affected by where the centre of rotation is (meaning the parallel axis theorum should not be used), yet I think he's wrong and intend to prove it.

Since I can't use any energies which are based on "I", I made 2 examples where I calculate the instantaneous kinetic energy of a rotating bar based on (1/2)mv^2 where v = wr. Please have a look at my work and tell me what you think. small letters imply variables in an equation, capital letters mean substituted values. For example, the formula may ask for m, but if the value for m is half the mass, the subbed in value will be M/2.

For the first case, I based the mass at the centre of mass. For the second question, I broke the piece into 2 separate pieces, which is why there are 2 centres of mass and the formula is multiplied by 2.

http://myfiles.dyndns.org/math/rotation_kinetic1.jpg

I tend to skip a lot of steps so if you don't understand why something was done, please ask.

http://myfiles.dyndns.org/math/rotation_kinetic2.jpg

On this one I calculated the different I values. One of them is just Ic, one of them includes the parallel axis theorum. For the examples, the one with the parallel axis of rotation has 4x as much energy. That same example has an I value 4x as big. Coincidence? HA!

So what do you think? Am I wrong or is my teacher wrong?

Last edited: Apr 1, 2004
2. Apr 1, 2004

### Staff: Mentor

I am not exactly sure what question your teacher asked. Could you please give the exact problem?

For a body rotating about a fixed axis with ω, the rotational KE will certainly depend on where that axis is. KE = 1/2 I ω2. Obviously I is axis-dependent. If your teacher is saying otherwise, he is incorrect. But let's see the exact statement of the question.

3. Apr 1, 2004

### ShawnD

Exact problem

http://myfiles.dyndns.org/math/rotation_problem1.jpg

It's rotating around the left side of the bar and not around the centre. That's why I used the parallel axis theorum to find "I". Then I used that "I" to find the rotational kinetic energy using the formula E = (1/2)Iw^2. The area where I calculated "I" using the parallel axis theorum is crossed out and his comment says "but we want Ic" and there's a little picture of a bar anchored on the left side with a dot in the middle labeled "c".

I go to one of those secondary schools with small classes. It was marked by the teacher, not a graduate student.

Can you tell me what your profession is? It would be great if I could say a professional engineer agrees with me (if I'm right).

Last edited: Apr 1, 2004
4. Apr 1, 2004

### Staff: Mentor

The total energy of the system is the sum of:
(1) spring potential energy
(2) gravitational PE of bar
(3) KE of bar

The KE of the bar at the moment shown is KE = 1/2 I ω2, where I = 1/3 M L2. I is taken about the pivot point, not the cm.

5. Apr 1, 2004

### ShawnD

That's another thing I found strange about the teacher's answer, the number of energies involved. The teacher had some weird term of E = (1/2)mv^2. I was thinking that wasn't true because after the bar rotates (or would) rotate 2 pi, it would be in the same position it was before. The centre of rotation is not moving.

My answer was going to include those 3 energies you listed but I never had enough time to finish. I finished the term for kinetic energy (which he marked as wrong), and I was half-way done the energy of the spring (it was a weird quadratic). I knew the gravity part was going to be a goofy integral so I left it to the end.
We had about 30 minutes to do this problem and it killed us all. Of the 19 students in that class, only 4 of them passed the test. Highest mark was less than 70%. I'm one of the 3 students who got 0.

This exam was a mid-term worth 15% of the year.

6. Apr 1, 2004

### Staff: Mentor

My PC crashed earlier before I had a chance to add this:

It is certainly OK to view the motion of the bar as a combination of rotation about the center of mass plus translation of the center of mass. With that view, the kinetic energy is:
KE = 1/2 Ic ω2 + 1/2MVc2

I assume this is what your teacher had in mind. I will leave it to you to show that this is exactly equivalent to my earlier expression for KE, where I treated the motion as being pure rotation about the fixed pivot point.