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Tangential Speed and Angular Momentum

  1. Jul 12, 2008 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations


    3. 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.
  2. jcsd
  3. Jul 12, 2008 #2


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    Isn't the mass of the skaters at the end of the rod and the rod itself you are treating as having 0 mass?

    Without doing the math your first part looks correct.
  4. Jul 12, 2008 #3
    Looking at the question again, I realize I misread it. It asks for the direction of the angular momentum before and after the skaters collide with the rod.
    I don't understand how there could be angular momentum before the skaters touch the rod, because the rod is not moving, so omega would equal 0.

    After the collision, wouldn't angular momentum be in the direction towards the axis of rotation?

    I'm really confused.
  5. Jul 12, 2008 #4
    The 1st part looks ok.
    i dont know too much about rotational inertia. but i saw 2 things that torubled me: the UNITS and the Direction. For your answer for B, you really dint give much of a direction and the units were off. When you did :
    " 1/3(60kg)(1.4m)^2*(omega)= 39.2 kg m*3.95/sec= 154 kg m/s" it should have ended up being 154kg m^2/s. When you square 1.4m, you also have to square the unit which is "m"
    think about it because your answer ends up not having the same units as a regular momentum units.
  6. Jul 12, 2008 #5
  7. Jul 12, 2008 #6
    I'm not an expert at rotational inertia, either (not even very good at it, really). But I do know units for L = I(omega) is kg*m^2/s

    It sounds to me as if there is a problem with the calculation of omega since it comes out as seconds and not rad/s

    and as far as i know omega = theta/t = (2pi*r/r)/t = 2pi/t
  8. Jul 12, 2008 #7


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    I agree with that.

    Now having calculated the new effective speed in part 1 and knowing the total combined mass of the 2 skaters, as well as their moment arm, shouldn't you have enough information to calculate the total angular momentum after they latch on?

    Just remember that moment arm is calculated from the center of mass. ("Light weight rods" should be your clue on that).
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