The combination of an applied force and a constant frictional forceproduces a co
ID: 1757049 • Letter: T
Question
The combination of an applied force and a constant frictional forceproduces a constant total torque of 35.5 N·m on a wheel rotating about a fixedaxis. The applied force acts for 5.95s. During this time the angular speed of the wheel increases from 0to 10.3 rad/s. The applied force isthen removed, and the wheel comes to rest in 59.6 s. (a) Find the moment of inertia of thewheel.kg·m2
(b) Find the magnitude of the frictional torque.
N·m
(c) Find the total number of revolutions of the wheel.
rev (a) Find the moment of inertia of thewheel.
kg·m2
(b) Find the magnitude of the frictional torque.
N·m
(c) Find the total number of revolutions of the wheel.
rev
Explanation / Answer
Given that the net torque is = 35.5 N.m Initial angular velocity is 1 = 0 Final angular velocity is 2 = 10.3 rad/s --------------------------------------------------------------- The torque is =I 35.5 N.m = I (2 - 1)/t I = 35.5 N.m *t / (2 - 1) =---------- kg.m2 where t = 5.95s Number of revalutions in thistime 1 = 1*t + ( 22 -12 )* t2 / 2 =---------- rad The frictional torque is = I* (3 - 2) /t =I* ( 0 - 2) /59.6s = ---------- N.m Where final angular velocity 3 = 0rad/s Number of revalutions in thistime 2 = 2*t + ( 32 -22 )* t2 / 2 =---------- rad Total angular diplacement is =1 + 2 =-------- rad Number of revalutions in thistime 2 = 2*t + ( 32 -22 )* t2 / 2 =---------- rad Total angular diplacement is =1 + 2 =-------- radRelated Questions
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