# Faraday's Dynamo & kinetic energy

1. May 2, 2014

### 6c 6f 76 65

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

(The picture is not from my textbook)

A Faraday Dynamo is used as a brake. The brushes are connected with an external resistance $R_\Omega$. The circular disk has mass m, and angular velocity $\omega_0$ when t = 0.

a) How long does it take until the kinetic energy of the disk is halved?

2. Relevant equations

$emf =\frac{B R^2 \omega}{2}$
$dW=P dt$
$P=\frac{(emf)^2}{R_\Omega}$

3. The attempt at a solution
I have no idea how to figure this problem out.
I've tried integrating dW from $\frac{1}{2}m \omega^2_0 R^2$ to $\frac{1}{4}m \omega^2_0 R^2$ setting that equal to the integral from t=0 to t of P dt (where I substitute P for the emf)

Looking at the answer, $t=\frac{m R_\Omega ln(2)}{B^2 R^2}$, I figured I need to integrate over $\frac{1}{x}$, in order to get the ln(2), but I've got no clue how to find that x. Any help much appriciated.

2. May 2, 2014

### TSny

Using energy ideas, how is the rate of decrease of the KE of the disk related to the electrical power produced?

3. May 3, 2014

### 6c 6f 76 65

Sorry, I don't quite know what you mean. The KE and the electrical power produced is linked via the angular velocity with the formula: $W=\frac{(\frac{B R^2 \omega}{2})^2}{R_\Omega}t=\Delta KE$

4. May 3, 2014

### TSny

Note that $\omega$ decreases with time. So, the electrical power is not going to be constant. That means the link between electrical power P and KE must be written in terms of an instantaneous rate of change of KE (with care given to signs).

5. May 3, 2014

### 6c 6f 76 65

$\frac{dKE}{dt}=-\frac{(\frac{B R^2 \omega}{2})^2}{R_\Omega}$

Last edited: May 3, 2014
6. May 3, 2014

### TSny

Good. What if you express $\omega^2$ on the right side in terms of KE and moment of inertia $\small I$? If you let K stand for KE, can you express the equation as follows? $$\frac{dK}{dt} = -b K$$
Here $b$ is a collection of constants.

7. May 5, 2014

### 6c 6f 76 65

Thank you so much! Here's what I got:

$\frac{dK}{dt}=-\frac{B^2 R^4 \omega^2}{4 R_\Omega}=-\frac{B^2 R^2}{2 m R_\Omega}K$, used that $K=\frac{1}{2}m \omega^2 R^2$
Solution for diff. equation: $K=K_0 e^{-\frac{B^2 R^2}{2 m R_\Omega}t}$
$\frac{1}{2}=\frac{K}{K_0}=e^{-\frac{B^2 R^2}{2 m R_\Omega}t}$
$ln(\frac{1}{2})=-ln(2)=-\frac{B^2 R^2}{2 m R_\Omega}t$
$t=\frac{2 m R_\Omega ln(2)}{B^2 R^2}$

But it's not quite right, there's a 2 here

8. May 5, 2014

### TSny

Check this. $K = \frac{1}{2}I\omega^2$. What is $I$ for a solid disk?

9. May 5, 2014

### 6c 6f 76 65

$I=\frac{1}{2}m R^2$, thank you so much!