Answer:
0.337 lb-in
Explanation:
From the law of conservation of angular momentum,
L' = L" where L = initial angular momentum of system and L" = final angular momentum of system
Now L = Iω + Mt where Iω = angular momentum of shaft + wheel and Mt = impulse on system due to couple M.
L' = Iω' + (-Mt) (since the moment about the shaft is negative-anticlockwise)
L' = Iω' - Mt where Iω' = angular momentum of shaft at t' = 0 + wheel and Mt = impulse on system due to couple M in time interval t = 70 s.
L" = Iω" where Iω" = angular momentum of shaft at t" = 70 s.
Now I = moment of inertia of system = mk² where m = mass of system = W/g where W = weight of system = 6 lb and g = acceleration due to gravity = 32 ft/s². So, m = W/g = 6lb/32 ft/s² = 0.1875 lb-s²/ft and k = radius of gyration = 2 in = 2/12 ft = 1/6 ft.
So, I = mk² = (0.1875 lb-s²/ft) × (1/6 ft)² = 0.00521 lb-ft-s², ω' = initial angular speed of system = 3600 rpm = 3600 × 2π/60 = 120π rad/s = 377 rad/s, ω" = final angular speed of system = 0 rad/s (since it stops), t' = 0 s, f" = 70 s and M = couple on system
So,
Iω' - Mt" = Iω"
Substituting the values of the variables into the equation, we have
Iω' - Mt" = Iω"
0.00521 lb-ft-s² × 377 rad/s - M × 70 s = 0.00521 lb-ft-s² × 0 rad/s"
0.00521 lb-ft-s² × 377 rad/s - 70M = 0
1.964 lb-ft-s = 70M
M = 1.964 lb-ft-s/70 s
M = 0.0281 lb-ft
M = 0.0281 lb × 12 in
M = 0.337 lb-in
