How Do Ice Skaters' Speeds Affect Their Rotation Upon Grabbing a Rod?

[smillphysics]

1. Two ice skaters, each with mass M = 72.5 kg, both moving with speed V = 9 m/s, approach each other along straight-line parallel paths that are separated by a distance of D = 3.6 m. When directly opposite each other, the skaters grab the ends of a light rod that is the same length as the distance between them. What is the initial speed of rotation of the joined skaters in rad/s?


2. Equations
maybe use conservation of momentum to find vfinal
m1vo+m2Vo=m1vf+m2vf
then use ac=v^2/r
r comes from d/2

3. this results in the incorrect answer. I'm unsure of how to solve this problem. Am I using the wrong equations?

 

[Doc Al]

While conservation of linear momentum holds, that's not what you need for this problem. Hint: Translate their linear speed into angular speed about the center. (If you like, you can think of something else that is conserved.)

 

[smillphysics]

so the v=9m/s which when converted into radians is 9*2pi= 56.55rad/sec,
then w=v/r,
so w= 56.55/3.6= 15.7
centripetal acceleration = r*w--> 3.6*15.7= 56.52m/s^2
then I'm not sure- are they an inelastic collision that is spinning? but I have acceleration so that doesn't help.

 

[Doc Al]

so the v=9m/s which when converted into radians is 9*2pi= 56.55rad/sec,

This isn't correct. Don't try to "convert" a linear speed into rad/s.

then w=v/r,

That's what you need. Hint: The radius ≠ 3.6 m. (Where's the center?)

 

[smillphysics]

w=9/1.8=5
then centripetal acceleration a=rw^2 = 5^2*1.8= 45m/s^2
then could I use ac=v^2/r to get the velocity? No that can't be correct. I'm highly confused.

 

[Doc Al]

w=9/1.8=5

Good.
ω = v/r = (9 m/s)/(1.8 m) = 5 rad/s
And you're done! No need to mess around with centripetal acceleration or any of that.

 

[smillphysics]

w- is the average velocity. Thanks for your help! Good thing someone enjoys physics.

 

[Doc Al]

w- is the average velocity.

ω is the rotational speed, not the average velocity.

Thanks for your help!

You're welcome.