Help On Swing Arm Velocity On Moving Body

[nattyb52]

I am trying to come up with the velocity of a swing arm mounted on the side of a moving body as described below. The main body moves linear for one half inch and is then brought to a stop and the arm is allowed to swing forward a total of 50 degrees. Does my solution outlined below make sense or am I missing something.
Any help is greatly appreciated.

What is the Average Velocity of the outer end of a 0.5m swing arm weighing 2.75 lbs (1.247 kg). with the weight evenly distributed the length of the arm. The arm extends horizontally at right angles to the moving body it is attached to, pivoting on a bearing at the inner end, with a small roller ball bearing holding up the outer end above a smooth flat surface. The inner end of the arm pivots on the edge of a 9 pound (4.082 kg) body, on linear bearings, that is accelerating at a rate of 533.33 ft/s (162.56 m/s) in a straight line for a total length of 1/2" (0.0127 m). When the body reaches the end of it's travel, and is brought to an abrupt stop, it is moving at a velocity of 2.02 m/s. The arm is held at an angle of 115.0 degrees (2.007 Rad) to the center line of the direction of travel of the body by a flange to prevent it from moving any further backwards. When the body stops, the arm is free to move
forward for a total of 50 degrees (0.873 Rad) towards the front of the body before it reaches it's own bump stop. There is a total external counter force to the forward motion of the arm of 51.80 Newtons.

What is the final equivalent linear velocity of the outer end of the arm and what is the average equivalent linear velocity?

Equations

Newtons Second Law: F=M*A
Newtons Second Law of Rotation: Torque=Iα
Moment of Inertia: I=1/3ML^2
Torque: τ=rFSin∅
Angular Velocity: ω2=ω0^2+2α∅

Solution:

Total body force F = 4.082 kg * 162.56 m/s = 667.2 Newtons

Total force on arm F = 1.254 kg * 162.56 m/s = 203.867 Newtons

Moment of Inertia of Arm around the end of the arm I = 1/3 * 1.254 kg * 0.5 m = 0.1045 kg m^2

Using the arm force minus the resistance means I have 152.067 N of force to create movement.
Torque of arm t = 0.5 m * (203.867 N - 51.800 N) * Sin 0.873 Rad = 68.909 N m

Torque = Iα So α = torque / I = 68.909 N m / 0.1045 kg m^2 = 659.369 rad/s^2

Max. Angular Velocity at end of swing = ω^2
= (2.02 m/s / 0.5 m)2 + (2 * 659.369 rad/s2 * 0.873 rad) = 34.1634 rad/s

Equivalent Max. Linear Velocity = 34.1634 rad/s * 0.5 m = 17.0817 m/s

Ave. Angular Velocity = ((2.02 m/s / 0.5 m) + 34.1634 rad/s) / 2 = 19.1017 rad/s

Equivalent Ave. Linear Velocity = 19.1017 rad/s * 0.5m = 9.55 m/s

[nattyb52]

I'm not an expert, can someone at least let me know if this is the right approach?

Thanks