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nos
nos is offline
#2
Feb5-13, 11:37 AM
P: 32
Yes, If you consider a mass being accelerated and rotates in a circle.
Then the acceleration is:

[itex]F=ma[/itex]
multiply both sides by r:

[itex]\tau=rma=r^{2}m\alpha[/itex]
where [itex]\alpha[/itex] is the angular acceleration.
Take this sum of all masses:

[itex]\sum r^{2}dm[/itex]

Or another way:

The force on a small element dm is:
[itex]dF=r\frac{d\omega}{dt}dm[/itex]
then the torque on this small mass dm is:
[itex]d\tau= rdF=r^{2}\frac{d\omega}{dt}dm[/itex]
integrating this over the total mass gives the total torque:
[itex]\tau=\int r^{2}dm\frac{d\omega}{dt}[/itex]

Hope it helps