The circular blade on a radial arm saw is turning at 277 rad/s at the instant the motor is turned off. In 22.0 s the speed of the blade is reduced to 85 rad/s. Assume the blade to be a uniform solid disk of radius 0.130 m and mass 0.400 kg. Find the net torque applied to the blade. Ugh?

Question

anonymous · Answer

Torque=Moment of inertia * alpha
alpha =change in omega /time=277-85/22
MI=mass*(radius)^2/2 multiply both to get the answer

anonymous · Answer

So 277-85/22=273.136=alpha
(.4*.13^2)/2=.00338=MI
.00338*273.136=.9232

.9232 is incorrect.  What am I still doing wrong?

jamesj · Answer

$$ \alpha = \frac{\Delta \omega}{\Delta t} = \frac{277-85}{22} = \ ...$$

jamesj · Answer

(actually negative of that.)

anonymous · Answer

(277-85)/22=8.727
Torque=-.0295