Question

1) The combination of an applied force and a friction force produces a constant total torque of 35.8 N · m on a wheel rotating about a fixed axis. The applied force acts for 5.90 s. During this time, the angular speed of the wheel increases from 0 to 10.1 rad/s. The applied force is then removed, and the wheel comes to rest in 60.1 s.

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

kg · m^{2}

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

(c) Find the total number of revolutions of the wheel during the
entire interval of 66.0 s.

2) A cylinder of mass 8.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 5.0 m/s.

(a) Determine the translational kinetic energy of its center of
mass.

J

(b) Determine the rotational kinetic energy about its center of
mass.

J

(c) Determine its total energy.

J

Answer #1

1)Solution:

a)

Angular acceleration α = (ω2- ω1)/t = 10.1/ 5.90 = 1.71
rad/s^2

I α = I *1.71 = 35.8

I = 35.8 / 1.71

= 20.93 kg m/s^2

b)

Here the wheel comes to rest due to the frictional torque τ
(friction)

τ (friction) = I α' = 20.93 α'

α' = - 10.1/60.1 = - 0.168 minus to show that the speed
reduces.

τ (friction) = - 20.93*0.168 = 3.52 N.m

==============================

c)

in time 5.8s

θ= 0.5 α t^2 = 0.5*1.71*5.90^2 = 29.76 radians.

Or from average angular velocity *time = (10.1/2)*5.90 =29.79
radians

In time 66.0 s

(9.5/2)*66.0 = 313.5 radians

Total angle traversed

313.5 +29.79 = 343.29 radians

343.29 / (2π) revolutions

= 54.66 revolutions

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