# Rigid body with fixed point

• I
Today I read a book in mechanics and encountered a funny proposition about rigid body with fixed point. Perhaps somebody will be interested to propose it to students as a task. This proposition is almost correct:)

Consider a rigid body with a fixed point ##O##. Let ##Oxyz## be a coordinate frame connected with this rigid body. Consider a unit sphere with center at ##O## as well. Now let us move the body from the initial position such that the axis $Oz$ describes a closed curve (without self-crossings) on the sphere and the projection of body's angular velocity on ##Oz## is equal to zero identically. It turns out that when the axis ##Oz## comes to the initial position other two axes will be rotated relative their initial position. The angle of rotation equals (up to the sign) the area of a figure drawn by the axis ##Oz## on the sphere.

Last edited:
vanhees71 and etotheipi

Arjan82
I don't know too much about mechanics, but how can an angle be equal to an area? Or am I not understanding it correctly?

unit sphere

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Sorry now I'm confused by units also.
The (origin-fixed) coordinate frame wanders about the unit circle such that the z axis draws a simple closed curve. Are you saying the inside wedge between the previous x-axis and the new x-axis has an area equal to the simple closed curve drawn on the sphere?

Once ##Oz## returns to its previous position, ##Ox## and ##Oy## must still define the same plane as at the start, but may be rotated from their starting positions within that plane. Meanwhile the area is that enclosed by the curve drawn on the sphere. For instance consider the simple case where ##\boldsymbol{\omega} = k \mathbf{e}_y## such that ##Oz## and ##Ox## rotate about the axis ##Oy##. When ##Oz## returns to its original position the curve bounds an area of a hemisphere, ##2\pi##, which is modulo ##2\pi## the same as the angle by which ##Ox## and ##Oy## are rotated from their original positions, ##0##.

I don't know how to prove the general case. Gauss-Bonnet theorem?

vanhees71, wrobel and hutchphd
I don't know how to prove the general case. Gauss-Bonnet theorem?
Oh no! It is simple. Just Euler angles and formula for the area in terms of integral along the boundary curve

Today I read a book in mechanics and encountered a funny proposition about rigid body with fixed point. Perhaps somebody will be interested to propose it to students as a task. This proposition is almost correct:)

Consider a rigid body with a fixed point ##O##. Let ##Oxyz## be a coordinate frame connected with this rigid body. Consider a unit sphere with center at ##O## as well. Now let us move the body from the initial position such that the axis $Oz$ describes a closed curve (without self-crossings) on the sphere and the projection of body's angular velocity on ##Oz## is equal to zero identically. It turns out that when the axis ##Oz## comes to the initial position other two axes will be rotated relative their initial position. The angle of rotation equals (up to the sign) the area of a figure drawn by the axis ##Oz## on the sphere.
What do you mean by the angle of rotation of the other axes?

Let's consider a simple 360° rotation around Ox. The area enclosed on the unit sphere is 2π. All body axes return to their initial orientations, so their relative rotation is zero. Of course Oy has traversed 2π, but Ox was fixed all the time.