Problem 2) What is the angular momentum of a 2.8 kg uniform cylindrical grinding wheel of radius 18 cm when rotating at 1500 rpm? b) How much torque is required to stop it in 7.0 s? 1500 rpm * 2pi rad = 9424.78 rad/m (9424.78 rad/m) /60 s = 157.08 rad/s I = 1/2mr2 I = 1/2(2.8 kg)(.18 m)2 = 0.04536 kgm2 L = Iw = (0.04536 kgm2)(157.08 rad/s) = 7.13 kgm2/s I have determined the angular momentum, but I am unsure of how to determine the torque required to stop it in 7 s How do you determine this with the torque formula and time involved? Any help?
The definition of torque is [tex] \mathbf\tau = \frac{d\mathbf{\mathrm L}}{dt} [/tex] If you want to change the angular momentum from your value to 0, then the average torque over 7 s most then be?
hmm....well you know the initial angular velocity ''157.08 rad/s''. I would say you need to calculate the angular deceleration of the spinning wheel from 157.08 r/s to 0 r/s in 7 secs using W(final)=W(initial)+a^t. I make the angular deceleration 22.44 rad s-2. Then using the formula T=I x a......cha ching !
w=w0+at 0=157.08+a(7) -157.08=7a a=-22.44 rad/s^2 since stopping T=Ia T=0.04536 kgm^2*-22.44 T=-1.02 m*N