Faraday's Dynamo & kinetic energy

AI Thread Summary
A Faraday Dynamo is analyzed as a braking mechanism, where the goal is to determine the time it takes for the kinetic energy of a rotating disk to halve. The relationship between the disk's kinetic energy and the electrical power generated is established through the equations for electromotive force (emf) and power. The discussion highlights the need to account for the varying angular velocity over time, leading to a differential equation that links the rate of change of kinetic energy to constants derived from the system's parameters. The solution involves integrating this relationship, ultimately leading to a formula for time that incorporates the mass, resistance, and magnetic field strength. The final expression for the time taken to halve the kinetic energy is confirmed to be t = (2 m R_Ω ln(2)) / (B^2 R^2).
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Homework Statement


hmpbasic.gif

(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.
 
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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?
 
TSny said:
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
 
6c 6f 76 65 said:
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).
 
\frac{dKE}{dt}=-\frac{(\frac{B R^2 \omega}{2})^2}{R_\Omega}
 
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6c 6f 76 65 said:
\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
 
6c 6f 76 65 said:
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?
 
I=\frac{1}{2}m R^2, thank you so much!
 
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