Faraday's Dynamo & kinetic energy

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Homework Statement

(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?

Homework Equations

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

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.

 

[TSny]

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

 

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

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

 

[TSny]

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

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).

 

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

 

[TSny]

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

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.

 

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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

 

[TSny]

used that K=\frac{1}{2}m \omega^2 R^2

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

 

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