Torque Required to Stop a Rotating Grinding Wheel

[needhelp83]

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?

 

[P3X-018]

The definition of torque is

$$\mathbf\tau = \frac{d\mathbf{\mathrm L}}{dt}$$

If you want to change the angular momentum from your value to 0, then the average torque over 7 s most then be?

 

[needhelp83]

t=(7.13 kgm^2/s)/7.0 s=1.02 kgm^2

Better?

 

[timmy]

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 !

 

[needhelp83]

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