# 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.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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 - f)/29.8

(9.6 + 2.4f) = 1.07(12 - f) = 12.84 - 1.07f

Solving for f we get: f = 0.93 lb

Substituting in (3) we get: α = (9.6 + 2.4 x 0.93)/31.8 = 0.37 rad/sec2

Initial angular velocity ω0 = 0

In two revolutions angular displacement θ = 4П

Applying the equation: ω2 - ω02 = 2αθ

ω2 - 0 = 2 x 0.37 x 4П = 9.29

ω = 3.05 rad/sec

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

h

Solution: Initial potential energy of slender rod AB = (PE)AB = (mg)h where h = height h (see fig)

(PE)AB = 15 x 1.5sin45O = 15.9 lb.ft

Similarly initial potential energy of slender rod BC = (PE)BC = 15.9 lb.ft

As the rods come down, the disc rolls to the right. At the instant angle θ = 0, both the rods are horizontal and the disc comes to rest (as it can no longer move further to the right). Hence, at θ = 0, kinetic energies (both translational and rotationsl) of the disc are zero.

Length of the spring in the initial configuration = 2 x 3 x cos45O = 4.24 ft

Length of the spring when θ is equal to 0 = 6 ft

Elongation of the spring = x = 6 - 4.24 = 1.76 ft

Potential energy stored in the spring at θ equal to zero = PESPRING = ½ kx2 = ½ x 4 x 1.762 = 6.2 lb.ft

Rotational KE of each rod at θ equal to zero = ½ Iω2

where I = moment of inertia of the rod about the axis passing through one of its ends = ML2/3

Mass of each rod = M = 15/32.2 = 0.47 slug

I = 0.47 x 32/3 = 1.41 slug ft2

Rotational KE of each rod at θ equal to zero = ½ x 1.41 x ω2

Rotational KE of both the rods at θ equal to zero = KEROT = 2 x ½ x 1.41 x ω2 = 1.41ω2

By the principle of conservation of mechanical energy:

Net initial PE of two rods = PESPRING + KEROT

2 x 15.9 = 6.2 + 1.41ω2

Solving we get ω = 4.26 rad/sec

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.

Solution: By Newton's second law, as applied to the rotational motion, applied torque is equal to the rate of change of angular momentum L.

τ = dL/dt where angular momentum L = Moment on inertia x Angular velocity = Iω

τdt = d(Iω) = Idω ........(1)

Initial torque acting on the bar = τ = 1000h Nm

Time duration of force dt = 0.03 sec

Moment of inertia of the rod about an axis passing through one end = I = ML2/3 = 8 x 22/3 = 10.67 kg.m2

Substituting values in (1): 1000h x 0.03 = 10.67dω

Change in angular speed dω = 1000h x 0.03/10.67 = 2.81h rad/sec ........(2)

Hence, initial angular speed can be taken as ω0 = 2.81h rad/sec

Initial rotational kinetic energy = ½ I ω02 = ½ x 10.67 x (2.81h)2 = 42.13h2 joule ............(3)

When the bar is horizontal its centre of gravity is raised by a distance of 1 m. Hence, increase in potential energy of the rod from vertical to horizontal position = mgh = 8 x 9.8 x 1 = 78.4 Joule

As the horizontal position is the final position before the bar begines to fall back, the initial KE must be equal to the increase in its PE in the horizontal position.

42.13h2 = 78.4

h = 1.36 m

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