# Rotational Kinetic Energy of Disks

1. Jun 19, 2013

### chrisa88

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

A disk with a mass of 4kg and a radius of 2m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at a constant rate of 20 J/rad; that is, the energy increases linearly with the angular displacement of the disk. If the disk starts from rest how many revolutions has it completed after 4s?

Question
I'm not exactly sure if I know how to solve this problem. I have solved problems that were similar, but the rotational kinetic energy of 20 J/rad is throwing me off. Why is the Krot= 20 J/rad?? I thought that Kinetic energy was in Joules only. My only thought is that this problem requires me to do the integral or derivative of the Krot=1/2Iω2 equation. Is my thought process wrong or am I interpreting something wrong here?
Help would be very much appreciated!!

Thank you!

-Christopher

2. Jun 19, 2013

### Curious3141

That 20J/rad is telling you that every radian of angle the disk turns through, it's gaining 20J of rotational kinetic energy.

This problem can be solved very easily by a repeated application of Chain Rule.

Let $K$ be the rotational kinetic energy. It's a variable that changes with time and angle.

$\theta$ is angle the disk has turned through (from the start), $\omega$ is the instantaneous angular speed ($\omega = \frac{d\theta}{dt}$) and $t$ is time.

You're given $\frac{dK}{d\theta}=20J/rad$.

What's the relationship between $\frac{dK}{dt}$, $\frac{dK}{d\theta}$ and $\frac{d\theta}{dt}$?

What's the relationship between $\frac{dK}{dt}$, $\frac{dK}{d\omega}$ and $\frac{d\omega}{dt}$?

With this, you should be able to set up a simple equation where $\omega$ cancels out. You're left with a simple number for $\frac{d\omega}{dt}$. From that, you should be able to determine the answer by simple integration (twice). Post again if you need help.

3. Jun 23, 2013

### chrisa88

Curious3141 I very much appreciate your reply!! That was very helpful. I believe I understand it now. I've been very busy with lab reports so I haven't been able to go back and make sure I understand it, but I will do that tomorrow. Again, thank you very much!!

-Christopher A.