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 · m²

(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