Recent content by kalbio

Ah I finally got the problem, I had to also take into consideration the acceleration of when the wheel was slowing down as well. Add the two accelerations and then use that for the Intertia. b) was just using the found intertia to solve for torque Thanks a bunch to those that helped!

Ah true...wouldn't the angular velocity and acceleration be this instead then 520 rev/min * 1min/60 sec = 8.67 rev/sec 8.67 rev /sec *2pi rads / rev = 54.454 rads /sec (This would be my actual angular velocity) So then my angular acceleration would be a = w/t a = 54.454 (rads/sec) / 23...

Uhm If I use the acceleration I got .1029 rads/sec^2 into the equation T = Inertia x angular acceleration That would be Intertia = Torque/angular acceleration Intertia = 44 / .1029 Intertia = 427.6 kg m^2 That didn't get me the right answer and it doesn't look like the right one. Is that...

Homework Statement A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +44 N·m is applied to the wheel for 23 s, giving the wheel an angular velocity of +520 rev/min. The external torque is then removed, and the wheel comes to rest...