Position #1:
radius, rā = 3 ft
Tangential speed, vā = 6 ft/s
By definition,Ā the angular speed is
Ļā = vā/rā = (3 ft/s) / (3 ft) = 1 rad/s
PositionĀ #2:
Radius, rā = 2 ft
By definition, the moment of inertia in positionsĀ 1 and 2 are respectively
Iā = (4 lb)*(3 ft)Ā² = 36 lb-ftĀ²
Iā = (4 lb)*(2 ft)Ā² = 16 lb-ftĀ²
Because momentum is conserved,
IāĻā = IāĻā
Therefore the angular velocity in position 2 is
Ļā = (Iā/Iā)Ļā
Ā Ā Ā = (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
vā = rāĻā = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
Fā = m*(vĀ²/rā) = (4)*(6Ā²/3) = 48 lbf
In position 2, the centripetal force is
Fā = (4)*(4.5Ā²/2) = 40.5 lbf
The radius diminishes atĀ aĀ rateĀ of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work doneĀ is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer:Ā 44.25 ft-lb
