Question

Problem 12.82

A 45kg figure skater is spinning on the toes of her skates at 1.1rev/s . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40kg , 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kgeach, 71cm long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 kg, 20-cm-diameter, 200-cm-tall cylinder.

Part A

What is her new rotation frequency, in revolutions per second?

Answer 1

Given,

Mass of the skater = Ms= 45 kg ; = 1.1 rev/sec

In the first case, Its hinted in the question that she can be viewed as a cylinder with following parameters:

Mc = 40 Kg ; diameter = 20 cm; radius = 10 cm = 0.1 m ; Length = L = 160cm = 1.6 meters

M(arms) = 2.5 kg (each) ; L' = 71 cm = 0.71 meters

We know that, the moment of inertia of cylinder is given by:

I(body )= 1/2 Mc r² = 1/2 x 40 x (0.1)² = 0.2 kg-m²

I' = I(rod) = 1/12 Mr L² = 1/12 x 2.5 x (0.71)² = 0.105 kg-m²

As the skater is rotating about perpendicular axis, her arms are not doing so, indeed their rotating is about center of body some distance away, that dist will be:

D = L/2 + r = 0.71/2 + 0.1 = 0.46 m

Now, we can apply the theorem of parallel axis to find the moment of inertia of her arms

I(arms) = I(center) + M(arms) x D² = I' + 2.5 x (0.46)² = 0.105 + 0.529 = 0.634 kg-m²

Total moment of inertis of the skater will be

I(total) = I(body) + 2 x I (arms)

I(total) = 0.2 + 2 x (0.634) = 1.47 kg-m²

Her Angular momentum will be,

L = I = 1.47 x 1.1 rev/sec

In the second case, when her arms are streched upwards

I(body)' = 1/2 m r² = 1/2 x 45 x (0.1)² = 0.255 kg-m²

From conservation of angular mometum,

L(intial) = L(final)

1.47 x 1.1 = 0.255 x '

' = 1.47 x 1.1 / 0.255 = 6.34 rev/sec

Hence, the new rot. freq = 6.34 rev/sec