Tangential Speed and Angular Momentum

[yellowgators]

Homework Statement

An ice-skater, with a mass of 60.0 kg, glides in a circle of radius 1.4 m with a tangential speed of 6.0 m/s. A second skater, with a mass of 50.0 kg, glides on the same circular path with a tangential speed of 5.0 m/s. At an instant of time, both skaters grab the ends of a lightweight, rigid set of rods, set at 90 degrees to each other, that can freely rotate about a pole, fixed in place on the ice. A) If each rod is 1.4 m long what is the tangential speed of the skaters after they grab the rods? B) What is the direction of the angular momentum of the before and after the skaters "collide" with the rods?

Homework Equations

A)Vtang=r(omega)
m1Vi1+m2Vi2=(m1+m2)Vf

B)L=I(omega)

The Attempt at a Solution

A) Vf=(m1mVi1+m2Vi2)/(m1+m2)=[60kg(6.0 m/s)+50kg(5.0m/s)]/(60kg+50kg)=5.54m/s
I didn't include the radius in my calculations because it remains the same for both skaters before and after the collision with the rods.
B)For the second part: I looked up the rotational inertia of a rod rotating at an axis located at its end: 1/3ML^2. Because the problem says lightweight rods, I think I need to set it equal to zero, but then I=0 and the angular acceleration would equal zero.
After the skaters collide with the rods:
omega is Vtang/r= (5.54m/s)/1.4m=3.95
1/3ML^2 for the first skater- 1/3(60kg)(1.4m)^2*(omega)= 39.2 kg m*3.95/sec= 154 kg m/s
for the second skater- 1/3(50kg)(1.4m)^2*(omega)=32.67 kg m*3.95/sec=129 kg m/s
154+129 kg m/s= 283 kg m/s

Is this right? Any help would be very much appreciated.

 

[dynamicsolo]

I'll start out by saying that I find this problem to be poorly written (I assume you're quoting it verbatim). It doesn't say whether the two skaters are moving on the path in the same or opposite directions and whether the direction(s) are clockwise or counter-clockwise are seen, say, from above. (I also find the description of the rod system unclear, but you have interpreted it in the same way I did...)

If they are going in the same direction, then I agree with your calculation

A) Vf=(m1mVi1+m2Vi2)/(m1+m2)=[60kg(6.0 m/s)+50kg(5.0m/s)]/(60kg+50kg)=5.54m/s
I didn't include the radius in my calculations because it remains the same for both skaters before and after the collision with the rods.

But an alternative acceptable answer, given the phrasing of the problem, would be

Vf = [ 60kg · (6.0 m/s) - 50kg · (5.0m/s) ] / ( 60 kg + 50 kg ) .

As for the second part, they are only asking for the direction of the angular momentum. Well, that would be perpendicular to the plane of the skaters' motion, and thus would point either straight up or straight down. But which direction it is depends on which way the skaters are circulating. The only thing that can be said for sure is that since the heavier skater is also moving faster, their angular momentum dominates in case the two are moving in opposite directions. So the total angular momentum points in the same direction as that of the larger skater, regardless of what the two are doing. If this skater is traveling counter-clockwise, the total angular momentum points upward (both before and after the collision); otherwise, it points downward.