Question

The combination of an applied force and a friction force produces a constant total torque of 35.9 N · m on a wheel rotating about a fixed axis. The applied force acts for 6.20 s. During this time, the angular speed of the wheel increases from 0 to 9.6 rad/s. The applied force is then removed, and the wheel comes to rest in 60.5 s.

(a) Find the moment of inertia of the wheel.

kg · m^{2}

(b) Find the magnitude of the torque due to friction.

N · m

(c) Find the total number of revolutions of the wheel during the
entire interval of 66.7 s.

revolutions

Answer #1

a)Angular acceleration α = (ω2- ω1)/t = (9.6-0)/ 6.2 = 1.55
rad/s^2

τ =I α

I *1.55 = 35.9

I = 35.9/ 1.55 = 23.16 kg m/s^2

b) The wheel comes to rest only due to the frictional torque τ
(friction)

τ (friction) = I α' = 23.16 α'

α' = - 9.6/60.5 = - 0.158 negative sign indicates that speed
reduces

τ (friction) = - 23.16*0.158 = 3.659 N.m

c) In time 6.2s

θ= 0.5 α t^2 = 0.5*1.55*6.2^2 = 29.79 radians.

Or from average angular velocity *time = (9.6/2)*6.2 =29.79
radians

In time 60.5 s

(9.6/2)*60.5 = 290.4 radians

Total angle traversed

290.4 +29.79 = 320.19 radians

320.19 / (2π) revolutions = 50.98 revolutions

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