# Rotational Motion

Please refer to the attachment for questions complete with diagrams.

1. The spool has a weight of 400 lb and a radius of gyration of kG = 1.6 ft. A horizontal force of P = 12 lb is applied to the cable wrapped around its inner core. If the spool is originally at rest, determine its angular velocity after the spool has turned 2 revolutions. The spool rolls without slipping and neglect the mass of the cable.

2. The assembly consists of two 15-lb slender rods and a 20-lb disk. If the spring is unstretched when theta = 45 degrees and the assembly is released from rest at this position, determine the angular velocity of rod AB at the instant theta = 0 degree. The disk rolls without slipping.

3. The suspended 8 kg slender bar is subjected to a horizontal impulsive force at B. The average value of the force is 1,000 N, and its duration is 0.03 s. If the force causes the bar to swing to the horizontal position before falling and coming to a stop, what is the distance h? Assume the bar is still vertical immediately after the impact time.

© BrainMass Inc. brainmass.com July 16, 2018, 12:50 pm ad1c9bdddf#### Solution Preview

Please refer to the attachment.

1. The spool has a weight of 400 lb and a radius of gyration of kG = 1.6 ft. A horizontal force of P = 12 lb is applied to the cable wrapped around its inner core. If the spool is originally at rest, determine its angular velocity after the spool has turned 2 revolutions. The spool rolls without slipping and neglect the mass of the cable.

Frictional force f

Solution: Let the frictional force acting on the spool be f.

Net force acting on the spool = F = P - f = 12 - f ............(1)

Mass of the spool = M = 400/32.2 = 12.42 slug

Translational acceleration a = Net force/Mass = (12 - f)/12.42 ft/s2 ...........(2)

Net torque acting on the spool = τ = 12 x 0.8 + f x 2.4 = (9.6 + 2.4f) lb ft

Moment of inertia of the spool = I = MkG2 = 12.42 x 1.62 = 31.8 slug ft2

Angular acceleration = α = τ/I = (9.6 + 2.4f)/31.8 .........(3)

For no slippage between the spool and the ground, acceleration a and angular acceleration α must satisfy the relation: α = a/R where r = Radius of the spool.

Hence, (9.6 + 2.4f)/31.8 = [(12 - f)/12.42] x (1/2.4)

(9.6 + 2.4f)/31.8 = (12 - ...

#### Solution Summary

Rotational motions diagrams are provided for slender rods.