Pg. 260 - 50, 56, 62, 66*,71, 74*, 76, 79*, 82, 86

50
I = SI_i =  
mass A + mass B + 2 rods + axis shift for rod 1 + axis shift for rod 2
= 5ml² +Ml²(1/6 + 1/4 + 9/4)
56
From table (page 249) inertia of a slab around a central axis is 1/12M(a² + b² )
From parallel axis theorem I = ¹/₁₂ (a² + b² ) + M(displacement of axis)² = ¹/₁₂M(a² + b² ) + 
62
70*9.8*0.4

71
F = 0.5t + 0.30t² so t = St = Fr = (0.5t + 0.30t²)*r
St = Ia so a = St / I = (0.5t + 0.30t²)*r /I a(3) = (0.5*3 + 0.3*9)*0.1 / 10^-3
w(t) = w( 0 ) + = 

76
a) Q = 1/2 a t² so a = 2Q /t² and w = 2Q /t
b) a = a*R
c) lower: Mg - T_L = Ma so T_L = M(g - a)
d) upper: T_U - f_k = Ma = M a R can be used to find f_k = T_U- M a R
during time t the mass M makes distance d = Q *R . Energy of system will be smaller by f_k Q R
Mg Q R - f_k Q R = 2*[1/2 M(2Q R /t)² ] + I(2Q /t)² ( two masses M move with speed v = w R and the pulley rotates )
finally: Mg Q R - (T_U- M a R)Q R = 2*[1/2 M(2Q R /t)² ] + I(2Q /t)²

T_U = [- 2*[1/2 M(2Q R /t)² ] - I(2Q /t)² + Mg Q R ] / Q R + M a R