Angular momentum of particle problem

[npiper7]

Homework Statement

A 3-lb collar can slide on a horizontal rod which is free to rotate about a vertical shaft. The collar is initially held at A= 0.5 ft by a cord attached to the shaft. A spring constant of 2 lb/ft is attached to the collar and to the shaft and is undeformed when the collar is at A=0.5 ft. As the rod rotates at the rate of (theta dot)= 16 rad/s, hte dord is cut and the collar moves out along the rod. Neglecting friction and the mass of the rod, determine (a) the radial and transverse components of the acceleration of the collar at A, (b) the acceleration of the collar relative to the rod at A, (c) the transverse component of the velocity of the collar at B = 1.5 ft.

Homework Equations

r' = transverse velocity r'' = transverse acceleration
(theta)' = radial velocity (theta)'' = radial acceleration

h=rvsin(phi)

The Attempt at a Solution

(a) Spring Force, Fs=kx=mar= 0
(mar = mass * transverse acceleration)

(b) m[ r'' - r (theta)'squared]

(c) h0=hf (initial = final)
r0v0sin(phi)=rvBsin(phi)
r(r0 theta') = rvB
vB= [r0(squared) * theta]/ rf

 

[Jhero]

(a) The radial component of the acceleration at A is 0, since the spring force is 0 and there is no friction or mass of the rod to cause any other acceleration. The transverse component of the acceleration at A is also 0, since the collar is initially at rest and there is no external force acting on it.

(b) The acceleration of the collar relative to the rod at A is equal to the acceleration of the collar itself, which is 0.

(c) To find the transverse component of the velocity at B, we need to use the conservation of energy equation, since the spring force is not constant due to the changing distance from the shaft.

Initial energy = final energy
Kinetic energy at A + Potential energy at A = Kinetic energy at B + Potential energy at B

At A:
Kinetic energy = 0, since the collar is initially at rest
Potential energy = 1/2 kx^2 = 1/2 * 2 * (0.5)^2 = 0.25 ft-lb

At B:
Kinetic energy = 1/2 mv^2 = 1/2 * 3 * v^2
Potential energy = 1/2 kx^2 = 1/2 * 2 * (1.5)^2 = 2.25 ft-lb

Setting the initial energy equal to the final energy:
0 + 0.25 = 1.5v^2 + 2.25
v^2 = 1
v = 1 ft/s

Therefore, the transverse component of the velocity at B is 1 ft/s.