|
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
|