Center of mass problem with rotation

[irishbob]

Homework Statement

An American football game has been canceled because of bad weather in Cleveland, and two retired players are sliding like children on a frictionless ice-covered parking lot. William 'Refrigerator' Perry, mass 162 kg, is gliding to the right at 7.41 m/s, and Doug Flutie, mass 81.0 kg, is gliding to the left at 11.0 m/s along the same line.

(c) The athletes had so much fun that they repeat the collision with the same original velocities, this time moving along parallel lines 1.06 m apart. At closest approach they lock arms and start rotating about their common center of mass. Model the men as particles and their arms as a cord that does not stretch. Find the velocity of their center of mass.
1.273 m/s <-- known

(d) Find their angular speed.

Homework Equations

w=|v|/|r|

The Attempt at a Solution

1.06m(2/3)=0.7067m
1.06/2=0.53m
0.7067m-0.53m=0.1767m
w=1.273 m/s / 0.1767m = 7.2043 rad/s

The problem is r is the center of mass, which is not the center of the line. I know that 2/3 of the total mass is 162kg and 1/3 is 81kg, and I know that's how you get r, but I don't know what to do with the ratio.

 

[tiny-tim]

welcome to pf!

top o' the mornin' to you, irishbob! welcome to pf!

(have an omega: ω )

… Find the velocity of their center of mass.
1.273 m/s <-- known

(d) Find their angular speed.
…
w=1.273 m/s / 0.1767m = 7.2043 rad/s

The problem is r is the center of mass, which is not the center of the line. I know that 2/3 of the total mass is 162kg and 1/3 is 81kg, and I know that's how you get r, but I don't know what to do with the ratio.

no, the c.o.m. velocity has nothing to do with the angular velocity

to find the angular velocity, you need the initial https://www.physicsforums.com/library.php?do=view_item&itemid=313" about the c.o.m. …

since angular momentum is always conserved in collisions (unlike energy), you know that will also be the final angular momentum about the c.o.m., and you can get the angular velocity from that