Physics+ Newton Laws, Problem 4: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 4

Problem

This problem is related to an effect on rotation of a figure skater when she pulls the hands closer to her body, thereby increasing the angular speed of rotation.

The physical problem that models this situation is as follows.
A point mass m initially rotates on a thread of length r₀ with initial angular speed ω₀.
We would like to shorten the radius of rotation and see how this is related to a speed of rotation.

To accomplish this, let's put the thread that holds a rotating mass through a thin tube positioned perpendicularly to a plane of rotation with its one end used as a center of rotation, so we can shorten the radius of rotation by pulling a thread from the other end of a tube.


Assuming the radius of rotation is changing with time and, finally, equals to r, what then will be the angular speed of rotation ω?

Solution

The solution is based on the Rotational (Angular) Momentum Conservation Law.
This law states that vector
L = r⨯ p
where r is a position vector relative to a center of rotation
and p is a momentum of motion of an object,
is preserved and does not change in direction or magnitude.

Consider the beginning and the end of the process of reducing the radius of rotation. In both these cases the motion of our object is rotational, and vectors of position r and velocity v are perpendicular to each other.
Therefore, their vector product has a magnitude
L = m·r·v
Since v=r·ω, L = m·r²·ω
This value is the same at the beginning and at the end of the process.
Based on the Law of Conservation of Angular Momentum, dependency of angular speed ω from radius r can be easily derived as
L = m·r²·ω = L₀ = m·r₀²·ω₀
r²·ω = r₀²·ω₀
ω = ω₀·r₀²/r²

Incidentally, the expression I=m·r² is called moment of inertia, so the Law of Conservation of Angular Momentum can be expressed as
L = I·ω = L₀ = I₀·ω₀

For example, if the radius is reduced by half, the angular speed will increase by a factor of four.

Let's consider a more practical case of a rotating figure skater, when she, to increase the speed of rotation, brings her arms close to a body.
This is a rotation of a solid object of variable geometry, and we will simplify our job by considering the following model.

Assume, there is a rotating object that has two parts.
One part, modeling arms initially stretched, but later brought tightly to the body, has mass m and rotates on initial radius r₀, which will be changed to a smaller radius r.
Another part, modeling the rest of a figure skater's body, has mass M and a constant radius of rotation R.

Assume, initial angular speed of rotation of a figure skater with arms stretched is ω₀.

The total angular momentum with arms stretched is
L₀ = m·r²₀·ω₀ + M·R²·ω₀.

Bringing the arms close to the body of a figure skater in our model will be reducing the radius of rotation of mass m from initial r₀ to final R which is smaller.
This will change the angular speed of rotation to ω.

The final angular momentum with arms close to the body and a new angular speed of rotation ω is.
L = (M+m)·R²·ω.

According to the Law of Conservation of Angular Momentum, the final angular momentum L is equal to the initial one L₀
L = L₀
Therefore,
m·r²₀·ω₀ + M·R²·ω₀ =
= (M+m)·R²·ω.
from which follows that the final angular momentum is
ω =
= ω₀·(M·R²+m·r²₀)/[(M+m)·R²]

Since r₀ (radius of rotation of stretched arms) is greater than R (radius of rotation of the rest of the body), the numerator of the above fraction is greater than denominator, which makes final angular momentum ω greater than the initial one ω₀.