A particle of mass m is constrained to move on a circle of radius R. This circle also rotates in space about a fixed point (P) on the circumference of the circle. The rotation of the circle is about an axis of rotation perpendicular to the plane of the circle and tangent to the circle, at point (P); the rotation is at a constant angular speed W.
In the absence of a gravitational force, show that the particle's motion about one end of a diameter passing through the pivot point and the center of the circle, perpendicular to the axis of rotation, is the same as that of a plane pendulum in a uniform gravitational field.
Explain why this result is reasonable.© BrainMass Inc. brainmass.com June 19, 2018, 1:02 am ad1c9bdddf
It is convenient in this type of problem to first write down the Lagrangian without constraints. Later on in the problem you put in the constraint in the form of a Lagrange multiplier term.
So, if the particle were completely free to move around then the appropriate Lagrangian would be just the kinetic energy:
L = 1/2 m v^2 (1)
Next we define co-rotating coordinates in which the circle is not moving. In those coordinates it's easy add the Lagrange multiplier terms to the Lagrangian. First we write (1) in polar coordinates with origin the pivot point P. This yields:
L = 1/2 m [r-dot^2 + r^2 theta-dot^2] (2)
Now let's define co-rotating coordinates in which the circle is not moving.
Points that are co-moving with the rotating circle will have constant r
and theta = alpha + omega t with constant alpha. Alpha and r can thus be taken as the co-rotating coordinates. Inserting theta = alpha + omega t in (2) gives:
L = 1/2 m [r-dot^2 + r^2 (alpha-dot + omega)^2] (3)
All we have done so far is a series of coordinate ...
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