You are designing a lathe motor, part of which consistsof a uniform cylinder who
ID: 1725303 • Letter: Y
Question
You are designing a lathe motor, part of which consistsof a uniform cylinder whose mass is 90kg and whose radius is .40 mthat is mounted so that it turns without friction on its axis,which is fixed. The cylinder is driven by a belt that wrapsaround its perimeter and exerts a constant torque. At t = 0,the cylinder's angular velocity is zero. At t = 25s, itsangular speed is 500 rev/min. (a) What is the magnitude of thecylinder's angular momentum at t = 25s? (b) At what rate is theangular momentum changing? (c) What is the magnitude of the torqueacting on the cylinder? (d) what is the magnitude of the frictionalforce acting on the rim of the cylinder?I tried doing part A by finding angular speed in terms ofrad/s and then using that velocity to find Angular momentum(L=m*v*r), but I'm coming up with the wrong answer. I'm notsure how to do the other parts either.
I tried doing part A by finding angular speed in terms ofrad/s and then using that velocity to find Angular momentum(L=m*v*r), but I'm coming up with the wrong answer. I'm notsure how to do the other parts either.
Explanation / Answer
Given that the mass of cylinder is m = 90 kg radius is r = 0.40 m At t =o initial angular velocity is 1 = 0rad/s At t = 25s initial angular velocity is 2 = 500 rev/min= 52.33 rad/s Change in time is t = 25s -------------------------------------------------------------------------- (a) The angular momentum at t = 25s is L2 = I*2 =(1/2)mr2*2 ( since momentof inertia of disk is I = (1/2)mr2 ) = 376.77 kg.m2/s (b) rate of change of angular momentum is L / t = ( L2- L1 ) / t = L2 / t ( since initial angularmomentum L1 =I*1 = 0 ) = 15.07 kg.m2/s2 (c) torque is = rate of change of momentum = 15.07 N.m
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