The combination of an applied force and a constant frictional forceproduces a co
ID: 1754018 • Letter: T
Question
The combination of an applied force and a constant frictional forceproduces a constant total torque of 35.0 N·m on a wheel rotating about a fixedaxis. The applied force acts for 5.94s. During this time the angular speed of the wheel increases from 0to 10.2 rad/s. The applied force isthen removed, and the wheel comes to rest in 59.8 s. (a) Find the moment of inertia of thewheel.1kg·m2
(b) Find the magnitude of the frictional torque.
2 N·m
(c) Find the total number of revolutions of the wheel.
3 rev (a) Find the moment of inertia of thewheel.
1kg·m2
(b) Find the magnitude of the frictional torque.
2 N·m
(c) Find the total number of revolutions of the wheel.
3 rev
Explanation / Answer
A)since the torque = I1 or I = /1 here 1 =(2-1)/t 2 =10.2rad/s 1 =0 t = 5.94s =35.0N calculate for I angular displacement is 1 =(22-12)/21 = ---------- rad = ---------- rad b)since the frictional torque =I2 here 2 =(3-2)/t here 2 =(3-2)/t 3 = 0 2 = 10.2 rad/s t = 59.8s calculate for angular displacement is 2 = =(32-32)/22 = ---------- rad = ---------- rad c) the final angular displacement is =1+2 = --------rad (we have 1rev = 2rad) = ----------rev = --------rad (we have 1rev = 2rad) = ----------revRelated Questions
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