# Flying beam-angular velocity

1. Nov 6, 2007

### amatol

A uniform beam with mass M = 168 kg and length L = 2.3 m slides broadside down along the ice at a speed of v0 = 8 m/s. A man of mass 85 kg, who is initially at rest, grabs one end of the beam as it goes past and hangs on as the beam and man go spinning down the ice. Assume frictionless motion. Assume zero moment of inertia for the man about a vertical axis through his center of mass. The moment of inertia of a uniform beam of mass M and length L about an axis through the center of the beam and perpendicular to it is ML2/12. Use the coordinate system shown in the picture, with the origin located at the initial position of the man and the z axis pointed out of the plane.

the coordinate system they show is pretty standard with y up and x horizontal. Beam travels left to right.
questions:
a) What is the y coordinate of the center of mass of the system before the collision?
pretty easy..got that one right. ycm = 1.15*168/(85+168) =.764m

b) What is the magnitude of the angular momentum of the man + beam system about its center of mass just before the collision?
Lcm = kg*m2/s

this is where I am very stuck. I tried treating it as a point particle and L=MRv based on L=I$$\varpi$$ I tried I as a rod using equation in question. I tried L=r X p, using r=length of rod - 2y$$_{}cm$$ (suggestion of the teacher, though I was confused on what exactly to do with that...I get that it would be the center of mass. the hints to the question are:
HELP: Only the beam is moving before the collision, and it is not rotating. Therefore, the angular momentum of the system is due entirely to the translational motion of the beam.
HELP: Remember that we want the angular momentum about the CM of the system. How far is the CM of the beam from the CM of the system?

there are a couple more parts to the problem... I think I can solve them if I can figure out b)

c) After the collision, what is the moment of inertia of the man + beam system about an axis perpendicular to the ice through the center of mass of the system?

d) After the collision, at what angular speed does the system rotate about its center of mass?

e) After the collision, what is the linear speed of the center of mass of the system?

thanks for any help!

2. Nov 6, 2007

### Astronuc

Staff Emeritus
Yes the beam is not rotating but translating, and it has zero angular momentum about it's CM. The beam's center of mass is the midpoint point at 1.15 m (assuming constant cross-section), so it is 1.15 m - 0.764 m from the CM after the collision. One could treat the beam as a mass at it's initial CM and apply that moment arm (1.15 m - 0.764 m).

3. Nov 6, 2007

### amatol

Thanks, but unfortunately this didn't really help me. This is for the angular momentum of the system before the collision, not after. I know the CM...it's the angular momentum I can't figure out. plugging in 1.15 m - 0.764 m by itself or as r doesn't seem to work. I am really at a loss how to solve this problem. wWhat formula should I use and where does the 1.15 m - 0.764 m fit in?? I've tried several variations of L=Iw and L=rXp.