## Tuesday, February 14, 2006

### Ice Skaters, Angular Momentum and Angular Acceleration

wo ice skaters, each of mass 80kg, are skating towards each other on parallel paths one meter apart each with a speed of 5m/s relative to the ground. As they pass each other, they reach out and link arms so that they remain 1m apart.
a) What is their center-of-mass velocity after they join hands?

b) What is their angular velocity?

c) Now, they pull inwards against each other so that the diameter of the circle they describe is 0.5m. Assuming that all their mass is located on the edge of this circle, what is their new angular velocity?

d) Now, once they reach this smaller separation, imagine a frictional force of 10N (directly opposing their velocity) acts on each skater. How long until they stop moving?

a) The joining of hands is exactly the same as any other collision – so linear momentum must be conserved. The initial linear momentum of the two skaters exactly cancels, so the momentum of the joined pair must also be zero. Thus the velocity of the center of mass of the joined pair is zero. They spin about the point directly between the two paths where they clasp hands.

b) There are a number of ways to correctly calculate the answer to this problem.

i. Observing that, if the skaters completed just one spin and then released hands, they would continue upon their original courses with their original velocities, it is clear that no energy can be lost in this collision – their initial linear kinetic energy is simply converted into rotational kinetic energy. Thus, we can see that the initial KE must be equal to the final rotational KE
1/2 Mv2 +1/2 Mv2 = 1/2 I ω2

I = 2Mr2 (= 2*80*(1/2)2 = 40kgm2)

Mv2=Mr2ω2

ii. Alternately, we can use conservation of angular momentum. The angular momentum of each of the two skaters about the axis of rotation is L = Mvrperp (where rperp is the perpendicular distance between the path of the skater and the axis - in this question rperp is equal to the radius of the circle the skaters follow). In this case, the initial, total, angular momentum is

L = 2Mvr.

By conservation of angular momentum, the final angular momentum of the spinning pair must also have this value.

Iω=2Mvr

iii. Finally, we can note that, in this situation, it is clear that the speed of the skaters is unchanged by the clasp of hands (this is fundamentally equivalent to saying that energy is conserved) then the speed of the skaters going around their circle is always 5m/s – thus: ω=v/r=10rad/s

c) While it is tempting to assume that the linear speed of the skaters remains unchanged as they pull together, we need to be careful here: At first glance it might appear that the extra inward pull that makes them come together is perpendicular to their motion and thus does no work (leaving their speed the same at the smaller radius), deeper thought reveals that their path, while the inward pull is in effect, is a spiral (this must be the case if they are to move to a different radius) – thus there is a component of their motion in the direction of the force. No matter how slow the spiral, the work done by the extra force is not something we can ignore! Thus it is incorrect to state that the speed at r=0.25m is the same as the speed at r=0.5m.

However, we can state that the angular momentum at r=0.25m and r=0.5m is constant – because the inward forces, acting through the center of rotation as they must, create no torques.

Li = Iiωi = Lf=Ifωf

As the final moment of inertia is 1/4 of the initial value (because I is proportional to the square of the radius of the circle), the angular velocity must increase by a factor of four to ensure that angular momentum is conserved

d) Now, we consider the effect of the frictional force on the skaters.

The torque due to the frictional force is given by τ=2*f*r (= 2*10*0.25= 5Nm)

Then, angular acceleration, α, is given by the rotational analogue of Newton’s 2nd law: τ=Iα

α =τ/I = 2fr/(2Mr2) = 0.5 rad/s2
thus, the time, t, till the skaters come to rest is given by: 0=ωf – αt

t=80s