In: Physics
A 45 kg figure skater is spinning on the toes of her skates at 1.0 rev/s . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg , 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 69 cm 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.
What is her new rotation frequency, in revolutions per second?
Conservation of Angular Momentum
L(final) = L(initial)
I * w (final) = I * w (initial)
Where I is the moment of inertia of the object
and w is the angular velocity of the object.
Case 1:
body is a cylinder spinning around its central axis: radius = 0.1 m, mass = 40kg.
The moment of inertia for a cylinder is:
I = m * r^2 / 2
I = 0.2 [kg * m^2]
sketer's arms are like rods: mass = 2.5kg, length = 0.69m
I = m * L^2 / 12
I = 0.0991875 [kg * m^2]
But sketers arms aren't rotating about their perpendicular axis, they are rotating about the center of sketer's body
which is some distance away. That distance can be expressed as half the length of her arm plus the radius of
sketer's body:
D = 0.69m / 2 + 0.1
D = 0.445 m
Using the parallel axis theorem, you can now find the moment of inertia of each arm:
I = I(center) + m * D^2
I = 0.0991875 + 2.5 * 0.445^2
I = 0.59425 [kg * m^2]
This is the moment of inertia for one arm. We already have the moment of inertia for sketer's body, so we can add
the body plus two arms to get sketer's total moment of inertia:
I = I(body) + 2 * I(arms)
I = 0.2 + 2 * (0.59425)
I = 1.3885 [kg * m^2]
So, using the equation for angular momentum, we know sketer's initial momentum is:
L(initial) = I * w
L = 1.3885 [kg * m^2] * 1 [rev/s]
Case 2:
skater is now one big cylinder:
radius = 0.1 m, mass = 45kg:
I = m * r^2 / 2 ==> I = 0.225 [kg * m^2].
And finally:
L(final) = L(initial)
I * w (final) = 1.3885
w = 1.3885 / I
w = 1.3885 / 0.225
w = 6.17 rev/s