Answers to homework problems:

Chapter 9 

15) Minimal lifting force is given by the condition of equilibrium:
     Sum of torques = 0

    a) left wheelbarrow
       525 N * 0.4 m + 60 N * 0.7 m = F * 1.3 m
       ===> F = 193.4 N

    b) right wheelbarrow
       525 N * 0 m + 60 N *0.7 m = F * 1.3 m
       ===> F = 32.3 N

30) Moment of inertia for a particle I = mr^2

   a) I(1) = 4.00 kg * (0.3 m)^2 = 0.36 kg*m^2
      I(2) = 10.0 kg * (0.7 m)^2 = 4.90 kg*m^2
      I(3) = 1.50 kg * (.96 m)^2 = 1.38 kg*m^2

   b) I(tot) = I(1) + I(2) + I(3) = 6.64 kg * m^2

   c) not true, moment of inertia depends quadratically on the lever arm

61) The change of angular speed is due to conservation of angular momentum.
    When the moment of inertia change (mass of the "people" changes its
    distance to the axis of rotation) then the angular speed has also to 
    change.
    Lets consider two cases:

    1) All pepople are at the outer surface of the station.
       The moment of inertia:
          I(1)=3*10^9 kg*m^2 + 500 * 70.0 kg * (82.5 m)^2 =
              =3.238*10^9 kg*m^2

    2) All pepople are at the center of the station i.e. at the axis of
       rotation. In this case the mass of the people does not add anything 
       to the moment of inertia:
          I(2)=3*10^9 kg*m^2

    Angular momentum conservation gives the relation:
     I(1) * omega(1) = I(2) * omega(2)
     ====> omega(2) = omega(1) *I(1)/I(2)

    Percentage change of the angular speed:
     [omega(1)-omega(2)]/omega(1) = [omega(1)-omega(1)*I(1)/I(2)]/omega(1)
                                  =   1-I(1)/I(2) = 1 - 1.0794 = -0.0794
    The station will increase its rotation by 7.94%