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Rigid Object Under Torque

  • Thread starter elkedoring
  • Start date
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?
 

Answers and Replies

18
0
the equation for torque is [tex]\tau=I\alpha[/tex], and for a disk rotating about its axis, [tex]I=\frac{1}{2}MR^2[/tex].

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

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