The combination of an applied force and a constant frictional forceproduces a co

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)                                                        = ----------rev

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