Solutions to Week 5 Assignment

1. (a) w = v/r = 0.50m/s/0.900m = 0.56rad/s.
   (b) 6.0m/2pr = 6.0m/2p(0.900m) = 1.1rev.
2. (a) The force of static friction between the inside wall of the rotor and the people's back keeps them from falling.
   (b) Use Newton's second law.
         SF_c = N = ma_c = mw²r
         SF_y = f - mg = 0
                   f = mg
                mN = mg
        mmw²r = g
               w = Ög/mr
                   = Ö9.8m/s²/(0.40)(2.5m)
                   = 3.1rad/s.
3. Let the x-axis point toward the center of curvature and the y-axis point upward. Use Newton's second law.
    SF_y = Ncosq - mg = 0
    SF_x = Nsinq  = ma_c = mv²/r
    Nsinq /Ncosq = mv²/r/mg
    tanq  = v²/rg
           q  = tan^-1(v²/rg)
               = tan^-1(26.8m/s)²/(122m)(9.8m/s²)
               = 31°
4. (a) v = 2pr/T = 2p(35,800km + 6371km)/(86,400s) = 3.07km/s.
        and v = 3.07km/s in the -y-direction at point C.
   (b)|v_av| = |Dr|/Dt = rÖ2/T/4 = 4rÖ2/T = 4(35,800km + 6371km)Ö2/(86,400s) = 2.76km/s, and v_av is in the same direction as Dr, which is 45° above the -x-axis. So v_av = 2.76km/s at 45° above the -x-axis.
   (c) a_av = Dv/Dt, so the average acceleration is in the same direction as Dv = v_B -v_A, which is 45° below the -x-axis.
        |Dv| = Ö[(Dv)_x]² + [(Dv)_y]² = Öv² + v² = vÖ2
        |Dv|/Dt = vÖ2/T/4 = 4vÖ2/T = 4Ö2(3.07 x10³m/s)/86,400s = 0.201m/s².
   So, a_av = 0.201m/s² at 45° below -x-axis.
   (d) The instantaneous acceleration at point D is in the +y-direction, since the acceleration is always directed radially inward for uniform circular motion. Its magnitude is a_c. Use Newton's second law and law of universal gravitation.
       SF_c =GmM_E/r² = ma_c
       a_c = GM_E/r²  = (6.673 x 10^-11N·m²/kg²)(5.975 x10²⁴kg)/(35,800 x 10³km + 6371 x10³km) = 0.224m/s².
   So, a = 0.224m/s² in the +y-direction. 
5. Dw = at
     a = Dw /t
         = (7200rpm)(2prad/rev)(1min/60s)/(2.0s) =190rad/s²
6. ^W = DFcosq  = TvDtcosq = (240N)(1.5m/s)(10.0s)cos30.0° = 3.1kJ.