Answers to homework problems:

Chapter 8 part 2.

48) tau = I * alpha  ---> alpha = tau/I
    For a disk the moment of inertia I = 1/2 mr^2
    alpha = tau/(1/2 mr^2) = 6.4 N*m /( 1/2 0.15 kg * 0.05^2 m^2)=34233 rad/s^2
56) if alpha is an angular acceleration, then
    R * alpha = a,                                 (1)
    where a is an acceleration which we need to find.
    The equation of motion for the weights:
    m2 * a = m2 * g - T2                           (2)
    m1 * a = T1 - m1 * g                           (3)
    and for the pulley
    {m * R^2 * alpha}/2 = (T2 - T1)*R - tau_f,     (4)
    where tau_f -- constant torque due to the friction, and m is a mass
    of the pulley. Solving the system of equations 1,2,3 and 4 we will get
    an answer:         
   
       a = [2(m2-m1)gR - tau_f]/[mR + 2(m2+m1)R] = 1.2 m/sec^2

88) Use formula 8.16 on the page 276 of the textbook:
       L = I * omega, 
    where I is a moment of inertia, and omega is an angular velocity.
    The direction of the L is in the direction of omega. 
    Since I = m * r^2 => 
       L = m * r^2 * omega = 8.5 * 10^4 kg*m^2/sec 
91) Using the same formula as for the problem above:
    L = I * omega, and omega = 2*pi/T, where T is a period of
    rotation, then,
        L = 2 * pi * I / T
    Rotation: I = 3/5 * M_moon * R_moon^2, where
    M_moon is a mass of the Moon, and R_moon is a Moon's radius.
        Lrot = 6/5 * pi * M_moon * R_moon^2 / T = 2.2 * 10^29 kg * m^2/sec
    Revolution: I = M_moon * R_orbit^2, where R_orbit is
    an orbit of the Moon.
       Lrev = 2 * pi * M_moon * R_orbit^2 / T = 2.6 * 10^34 kg*m^2/sec  

Chapter 9 part 1

 8) Using the formula 9.1 from the page 318, 
    Stress = F/A = 150N / (0.2 m * 0.3 m) = 2500 N/m^2    

11) Using the formula 9.4 from the page 318,
    Y = [F/A] / [deltaL/L_initial]
    In our case delatL = L_final - L_initial = 1.3026 - 1.3 = 0.0026 m
    F/A = 600 N / (pi * (0.002m)^2/4) = 1.2 * 10^7 N/m^2
    So, finally,
            Y = 1.2 * 10^7/ [2.6 * 10^(-3)/1.3] = 9.6 * 10^10 N/m^2
      
32) Weight of the athlete is W = m * g = 75kg * 9.8 m/sec^2 = 735N
    Pressure is P = W/A = 735N/0.0125m^2 = 5.9 * 10^4 Pa


33) Pressure = F * cos(alpha) / A,
    where F is a weight of the skier, alpha is an angle of inclination
    and the A is an area of the contact of skis, so
    Pressure = 90kg * 9.8 m/sec^2 * cos15 / 0.4 m^2 = 2.12 * 10^3 Pa