The equations governing the motion of a ball of mass m, radius R rolling on a table rotating at constant angular velocity Ω which are derived using Newton's laws are: (I present these for comparison) (I+mR2)˙ωx=(I+2mR2)Ωωy−mRΩ2y(I+mR2)˙ωy=−(I+2mR2)Ωωx+mRΩ2x˙x=Rωy˙y=−Rωx
Where x,y,ωx,ωy are absolute values measured in the rotating frame (x,y being positions and ωx,ωy angular velocities of the ball). To express the position, velocity, etc in the inertial XYZ frame we can perform a change of variables: X=xcosθ−ysinθY=xsinθ+ycosθ
The equations written above are correct as far as I know. Now I've tried to derive these equations using the Lagrangian approach, but my equations differ slightly from the above. I'll share my work here:
We start with the standard formulation of the Lagrangian equations of motion: ddt(∂L∂˙qi)−∂L∂qi=Qi
For this system, there are no non-conservative forces doing work at any time so Qi=0 (Assuming no-slip here). Now the kinetic energy of the system is: T=12m‖v‖2+12I‖ω‖2
Where v,ω are the absolute linear and angular velocities of the ball, I is the moment of inertia of the center of mass. I'll proceed in the rotating frame basis xyz, where the position r of the ball is given by →r=xˆi+yˆj
Using velocity kinematics in rotating frames, the absolute velocity of the ball is given by: →v=→Ω×→r+→vxyz
Where →vxyz is the velocity of the ball in the rotating frame: →vxyz=˙xˆi+˙yˆj
So the absolute velocity →v after expanding gives: →v=(˙x−Ωy)ˆi+(˙y+Ωx)ˆj
It follows from taking the magnitude of the absolute velocity: ‖v‖2=(˙x−Ωy)2+(˙y+Ωx)2
The first unknown term in the Lagrangian equations of motion. The absolute angular velocity →ω is more straight-forward: →ω=→Ω+→ωxyz
Where →ωxyz is the angular velocity of the ball in the rotating frame, where one can show it DOES NOT have a component in the ˆk direction if we impose no-slip kinematics (which we are), so: →ωxyz=ωxˆi+ωyˆj
And since →Ω=Ωˆk
The absolute angular velocity is: →ω=ωxˆi+ωyˆj+Ωˆk
Then it follows that: ‖ω‖2=ω2x+ω2y+Ω2
The potential energy of the system is constant and doesn't affect the equations of motion so, the Lagrangian becomes: L=12m[(˙x−Ωy)2+(˙y+Ωx)2]+12I[ω2x+ω2y+Ω2]
Now the constraint equations are the no-slip conditions: →vxyz=→ωxyz×→R
Where →R=Rˆk of course, then we have two conditions: ˙x=Rωy˙y=−Rωx
Which are non-holomonic constraints, but given their simple nature I opted out of doing Lagrangian multipliers and simply substituted them into the equations of motion (I reworked it with Lagrangian multipliers after and got the same thing). Now after substituting the constraints, the Lagrangian becomes:
L=12(m+IR2)(˙x2+˙y2)−mΩ˙xy+mΩ˙yx+12mΩ2(x2+y2)+12IΩ2
From here, applying the Lagrangian equation above with Qi=0 I get the following equations of motion: (m+IR2)¨x−2mΩ˙y−mΩ2x=0
(m+IR2)¨y−2mΩ˙x+mΩ2y=0
And using the no-slip conditions again we can rewrite: (I+mR2)˙ωx=2mR2Ωωy−mRΩ2y(I+mR2)˙ωy=−2mR2Ωωx+mRΩ2x
Now if you compare these last two equations with the ones I wrote in the beginning, the only difference is in the first term on the right-hand side. Look at these two for instance: (I+mR2)˙ωx=(I+2mR2)Ωωy−mRΩ2y(I+mR2)˙ωx=2mR2Ωωy−mRΩ2y
The ONLY difference is the missing I term! I'm missing the moment of inertia for some reason, why is that? What's wrong about my Lagrangian approach?
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