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Parabolic.

Short Answer

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au=u¨u2+v2+uu2+v2u˙2+2vu2+v2u˙v˙+-uu2+v2v˙2-uu2+v2ϕ˙2av=v¨u2+v2-uu2+v2u˙2+2uu2+v2u˙v˙+-vu2+v2v˙2-uu2+v2ϕ˙2az=ϕ¨uv+2vӬ˙ϕ˙+2vvϕ˙

Step by step solution

01

Given Information

The Elliptical cylinder.

02

Definition of cylindrical coordinates.

The coordinate system primarily utilized in three-dimensional systems is the cylindrical coordinates of the system. The cylindrical coordinate system is used to find the surface area in three-dimensional space.

03

Find the value.

The velocity and scale factors are given below.

hu=u2+v2hv=u2+v2hÏ•=uvrË™=uË™u2+v2eÁåœu+vË™u2+v2eÁåœv+Ï•uveÁåœÏ•

The lagrangian in the parabolic coordinates are given below.

L=12mr˙2-Vu,v,ϕL=12mu2+v2u˙2+u2+v2v˙2+u2v2ϕ2-Vu,v,ϕ

Apply Euler-lagrange equation, the above equation becomes as follows.

ddt∂L∂u˙=mu2+v2u¨+uu˙2+2vvu˙=mv˙2+ϕ˙2u-∂V∂u

Solve further.

mu¨u2+v2+uu2+v2u˙2+2vu2+v2u˙v˙+muu2+v2v˙-uu2+v2ϕ˙2=-1h2∂V∂umv¨u2+v2-uu2+v2u˙2+2vu2+v2u˙v˙+vu2+v2v˙2-vu2+v2ϕ˙2=-1hv∂V∂umϕ¨uv+2vu˙ϕ˙+2vv˙ϕ˙=-1h2∂V∂ϕ

Divide the above equation by respective scalar factors.

au=u¨u2+v2+uu2+v2u˙2+2vu2+v2u˙v˙+uu2+v2v˙-uu2+v2ϕ˙2av=v¨u2+v2-uu2+v2u˙2+2uu2+v2u˙v˙+vu2+v2v˙2-vu2+v2ϕ˙2az=ϕ¨uv+2vu˙ϕ˙+2vv˙ϕ˙

au=u¨u2+v2+uu2+v2u˙2+2vu2+v2u˙v˙+uu2+v2v˙2av=v¨u2+v2-uu2+v2u˙2+2uu2+v2u˙v˙+vu2+v2v˙2az=z¨

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Most popular questions from this chapter

Generalize Problem 3 to see that the direct product of any two isotropic tensors (or a direct product contracted) is an isotropic tensor. For example show thatϵijkϵlmnis an isotropic tensor (what is its rank?) andϵijkϵlmnδjnis an isotropic tensor (what is its rank?).

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In equation(5.12) , find whetherA×(B×C) is a vector or a pseudovector assuming

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Hint: Count up the number of det A factors from pseudovectors and cross products.

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