# I Lagrangian at a rolling sphere

1. Oct 23, 2016

### sgh1324

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 cant 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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2. Oct 23, 2016

### Staff: Mentor

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 $KE=0.5 m v^2$