Question

17. 53 A small grinding wheel is attached to the shaft of an electric motor that has a rated speed of 3600 rpm. When the power is turned off, the unit coasts to rest in 70 s. The grinding wheel and rotor have a combined weight of 6 lb and a combined radius of gyration of 2 in. Determine the average magnitude of the couple due to kinetic friction in the bearings of the motor.

Answer 1

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