- 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?

** Answer:**

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 Mv

^{2}+1/2 Mv

^{2}= 1/2 I ω

^{2}

I = 2Mr^{2} (= 2*80*(1/2)^{2} = 40kgm^{2})

Mv^{2}=Mr^{2}ω^{2}

ω=v/r = 10 rad/s

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 = Mvr_{perp} (where r_{perp} is the perpendicular distance between the path of the skater and the axis - in this question r_{perp} 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

ω=v/r=10 rad/s

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

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.

- L

_{i}= I

_{i}ω

_{i}= L

_{f}=I

_{f}ω

_{f}

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

- ω

_{f}=40 rad/s

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/(2Mr

^{2}) = 0.5 rad/s

^{2}

_{f}– αt

t=80s

## No comments:

Post a Comment