Magnetostatics - Rotation of circular disc

AI Thread Summary
A rotating disc with a radius of 3m and mass of 10^4 kg is analyzed under a magnetic flux density of 0.5T. The disc, initially spinning at 3000 revolutions per minute, experiences a sudden load of 10^-3 ohms connected between its rim and axis. The key equation for the electromotive force (emf) produced is w*a^2*B/2, which is used to relate the disc's kinetic energy to the energy loss due to the load. The discussion clarifies that the loss in kinetic energy should equal (V^2/R)*T, leading to the determination that the time to slow to half its initial speed is T=6.16 seconds. The participants emphasize the importance of correctly equating the energy loss to find the solution.
Keano16
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Homework Statement




A disc has radius a and rotates with angular frequency w. Magnetic flux density is B. Such a disc of mass 10^4 kg and radius 3m is rotating freely at 3000 revs/min in a field of 0.5T. A load of 10^-3 ohms is connected suddenly between the rim and the axis of the disc. How long would it take the disc to slow to half its initial speed?


Homework Equations





The Attempt at a Solution



I found an equation for the emf produced that is: w*a^2*B/2 . Using this I wanted to equate two equations -- kinetic energy of the rotating disc at half the intiial speed = (V^2/R)*T.

The answer is meant to be T=6.16s, but that's not what I have.

Thanks for any help.
 
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Keano16 said:
I found an equation for the emf produced that is: w*a^2*B/2 . Using this I wanted to equate two equations -- kinetic energy of the rotating disc at half the intiial speed = (V^2/R)*T.

No, the loss in kinetic energy should be equal to (V^2/R)*T.
 
Ah I see now -- thanks for your help!
 

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