Torque needed to stop a grinding wheel in 6.5 s

[AstroturfHead]

This is a problem I'm having a world of problem with:

A. What is the angular momentum of a 2.56 kg uniform cylindrical grinding wheel of radius 17.5 cm when rotating at 1480 rpm?

Correct: 6.08 kg*m^2/s

B. How much torque is required to stop it in 6.50 s?

So after 5 tries, I get that Torque = Moment of Inertia * Alpha.

Alpha = Delta (w)/ Delta (t)
Then I = 1/2*m*r^2
and I get .9347 N*m, but the computer does not. Any suggestions as to what I am doing wrong?

 

[dextercioby]

U've given it 4 sign.digits cf.3 before...

If A is correct (manipulating units),then B is correct,as well.I don't c what it could be.

Daniel.

 

[AstroturfHead]

Well I have lot of these homework questions. Heres another one that I can't figure out.

An Atwood machine consists of two masses, m1 = 6.80 kg and m2 = 8.55 kg, connected by a cord that passes over a pulley free to rotate about a fixed axis

Then there's a picture but I don't know how to link to it. Its just 2 weights on the end of a string around a pully.

The pulley is a solid cylinder of radius R0 = 0.535 m and mass 0.771 kg. Determine the acceleration a of both masses. Ignore friction in the pulley bearing.

10. [1pt]
What percentage error in a would be made if the moment of inertia of the pulley were ignored? Do not enter units.

Any ideas on there either? I did the same kinda thing and said T = F*R=I*Alpha etc etc