Lagrangian at a rolling sphere

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SUMMARY

The discussion centers on the Lagrangian mechanics of a rolling sphere on an inclined wall at an angle θ. The derived Lagrangian is L = 7/10 mv^2 + mgx sin(θ), where p represents the conjugate momentum, leading to the equation 7/5 mv = p. The confusion arises from the factor of 7/5 in the momentum equation, which is clarified as a result of the relationship between linear and angular momentum in Lagrangian mechanics, rather than traditional Newtonian momentum.

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  • Proficiency in calculus, particularly partial derivatives
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sgh1324
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A sphere is rolling inclined wall (θ radian).
and the momentum of that sphere is

L = 1/2 mv^2 + 1/5 mv^2 + mgx sin(θ) = 7/10 m v^2 + mgx sin(θ)
∂L/∂v = p
7/5 m v = p

but I can't understand why the factor of mv is 7/5.

p is the linear momentum of sphere.

which means the factor of mv must be always 1 (my thought)

p = mv

↑that was an unchanging truth of my physical world.

why rolling effect to the momentum? angular momentum of the sphere has no relation with p direction!
 

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sgh1324 said:
∂L/∂v = p
7/5 m v = p

but I can't understand why the factor of mv is 7/5.

p is the linear momentum of sphere.
Well, p isn't really the momentum from Newtonian analysis. It is the conjugate momentum from Lagrangian analysis. They have similar names, but they are only equal if ##KE=0.5 m v^2##
 
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