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

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SUMMARY

The discussion focuses on calculating the time and number of revolutions for a grinding wheel, modeled as a uniform solid disk with a radius of 7 cm and mass of 2 kg, under a constant torque of 0.6 Nm. The correct approach involves using the torque equation τ = Iα, where I = 1/2 MR² for a disk. The participant initially calculated the circular acceleration incorrectly, leading to an overestimation of time, which should be 1.03 seconds to reach 1200 revolutions per minute, not 2.06 seconds. The final number of revolutions can be derived from the correct time and acceleration values.

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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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