Q6.14 Skaters (K)

Two skaters, each of mass 50 kg, approach each other along parallel paths separated by 2.9 m. They have equal and opposite velocities of 1.4 ms-1. The first skater holds a light pole, 2.9 m long, at one end; the second skater grabs the other end of the pole as it passes him.
  1. What conservation laws are applicable here? Explain carefully.
  2. Assuming that the ice is frictionless, describe the motion of the two skaters after they are connected by the pole.
  3. By pulling on the pole, the skaters reduce their separation to 0.94 m. What is their angular speed then?
  4. Calculate the kinetic energy of the skaters before and after they reduce their separation. Explain the energy difference.
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Hint

If there are no net external forces, what quantities are conserved?
 

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Solution

Graphic - No title - angmomcons1
  1. Linear momentum is conserved because the total external force is zero – horizontal frictional forces on the ice can be neglected, and the vertical gravitational forces downwards are cancelled by equal reaction forces upwards. Note that the forces the skaters exert on one another are internal: they cannot change the total linear momentum of the system, even though they do change the linear momentum of the skaters individually.

    Angular momentum (about any origin) is conserved, because the total external torque (about any origin) is zero – each mass is acted on by a gravitational force and a reaction force, which are equal and opposite and have the same line of action, so the torques cancel out for any origin.

  2. After the skaters are linked by the pole, the system consisting of the two skaters rotates about its centre of mass (CM). The total linear momentum is zero, so the centre of mass remains at rest. (In fact, the problem is viewed in the centre of mass frame of reference.)

    The total angular momentum about CM before the linking was L1=mud, in a direction out of the diagram plane. After the linking, the total angular momentum is L2=mvd. Since L1=L2, v=u. Thus the skaters continue to move with their original speed, but now in a circle, centred on the CM.

  3. The forces exerted by the skaters on each other are internal to the system, and cannot generate an external torque. The total angular momentum thus remains constant as the skaters reduce their separation by pulling on the pole. When the separation is d=2.9 m and v=1.4 m s-1 

    L=mvd=203 kg m2 s-1

    When the separation is reduced to d=0.94 m,

    L=mvd=203 kg m2 s-1
    Hence
    v=4.3 m s-1
    The final angular velocity is
    ω=v/r=2v/d=9.1 rad s-1

  4. Before reduction the (rotational) kinetic energy is

    \[ K_i=2\times\frac{1}{2}mv^{2}=98\,J. \]
    After reduction the kinetic energy is
    \[ K_f=2\times\frac{1}{2}m(v^{\prime})^{2}=925 \,J \]

    The additional energy comes from the work done by the internal forces, converting biochemical energy stored in the muscles of the skaters into mechanical energy.