Rigid Object Under Torque: Solving for Time and Number of Revolutions

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A grinding wheel, modeled as a uniform solid disk with a radius of 7 cm and mass of 2 kg, accelerates under a constant torque of 0.6 Nm. The calculations for circular acceleration were initially incorrect, leading to an overestimation of the time to reach the final speed of 1200 rev/min. The correct approach involves using the moment of inertia for a disk, which is I=1/2 MR^2, to accurately calculate the angular acceleration. After correcting the calculations, the time to reach operational speed is determined to be 1.03 seconds. The discussion highlights the importance of applying the correct formulas for torque and angular motion.
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A grinding wheel is in the form of a uniform solid disk of radius 7cm and mass 2kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.6 Nm the motor exerts on the wheel. a) How long does the wheel take to reach its final operational speed of 1200 rev/min? b) Through how many revoulutions does it turn?



Torque= (m*r^2)*circular accel.
circular veloc. Final= circ.veloc. Intial + circular accel.*time



Ok, so I tried to use the above torque equation to solve for circular acceleration. So, .6/(2*.07^2) = circular accel.= 61.22 rad/s. Then I tried putting that value back into the other equation. So, 125.66 rad/s(I converted it)-0=61.22*t. Thus, t=2.06s. However, the time I am supposed to get is 1.03s (half of what I got). Could someone please explain where I went wrong?
 
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the equation for torque is \tau=I\alpha, and for a disk rotating about its axis, I=\frac{1}{2}MR^2.

i think that should solve your problem. cheers
 
Thank you that solved the problem!
 

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