Lagrangian for a rolling disk on horizontal plane

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For a disk rolling on a horizontal plane, the total kinetic energy is the sum of the translational kinetic energy of the center of mass and the rotational kinetic energy about the center of mass, expressed as T = 1/2 (M V^2) + 1/2(I ω^2). Some texts simplify the problem by only considering the rotational kinetic energy, T = 1/2 I ω^2, neglecting the translational component. This approach can lead to confusion regarding the complete dynamics of the system. The discussion highlights the importance of including both kinetic energy terms for accurate analysis. Understanding this distinction is crucial for correctly applying Lagrangian mechanics to rolling objects.
sayf alawneh
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for a disck rolling on a horizontal plane the kinetic energy should be the kinetic energy of the CM of the disk with respect to the origin plus the kinetic energy due to the rotation of the disc about his CM
so T= 1/2 (M V^2) +1/2(I ω^2)
where M is the mass of the disk and V is the velocity of the CM and I is the moment of inertia of the disk and W is the angular velocity about the CM

am i wrong ?
why some books sole this problem for rotational kinetic energy only and ignore the term of kinetic energy that depends on the velocity of the center of mass with respect to the origin
in ohter words they consider T = 1/2 I ω^2 only :(
 
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You would normally have to consider the translational motion also. In which books did you find this so that I can give a look at the problem.
 
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