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Physics LibreTexts

13.5: Acceleration Components

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In Section 3.4 of the Celestial Mechanics “book”, I derived the radial and transverse components of velocity and acceleration in two-dimensional coordinates. The radial and transverse velocity components are fairly obvious and scarcely need derivation; they are just ˙ρ and ρ˙ϕ. For the acceleration components I reproduce here an extract from that chapter:

“The radial and transverse components of acceleration are therefore (¨ρρ˙ϕ2) and (ρ¨ϕ+2˙ρ˙ϕ) respectively.”

I also derived the radial, meridional and azimuthal components of velocity and acceleration in three-dimensional spherical coordinates. Again the velocity components are rather obvious; they are ˙r,r˙θ and rsinθ˙ϕ while for the acceleration components I reproduce here the relevant extract from that chapter.

“On gathering together the coefficients of ˆr,ˆθ,ˆϕ we find that the components of acceleration are:

  • Radial: ¨rr˙θ2rsin2θ˙ϕ2
  • Meridional: r¨θ+2˙r˙θrsinθcosθ˙ϕ2
  • Azimuthal: 2˙r˙ϕsinθ+2r˙θ˙ϕcosθ+rsinθ¨ϕ. "

You might like to look back at these derivations now. However, I am now going to derive them by a different method, using Lagrange’s equation of motion. You can decide for yourself which you prefer.

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We’ll start in two dimensions. Let R and S be the radial and transverse components of a force acting on a particle. (“Radial” means in the direction of increasing ρ; “transverse” means in the direction of increasing ϕ.) If the radial coordinate were to increase by δρ, the work done by the force would be just Rδρ. Thus the generalized force associated with the coordinate ρ is just Pρ=R. If the azimuthal angle were to increase by δϕ, the work done by the force would be Sρδϕ. Thus the generalized force associated with the coordinate ϕ is Pϕ=Sρ. Now we do not have to think about how to start; in Lagrangian mechanics, the first line is always “T= ...”, and I hope you’ll agree that

T=12m(˙ρ2+ρ2˙ϕ2).

If you now apply Equation 13.4.12 in turn to the coordinates ρ and ϕ, you obtain

Pρ=m(¨ρρ˙ϕ2)andPϕ=mρ(ρ¨ϕ+2˙ρ˙ϕ),

and so

R=m(¨ρρ˙ϕ2)andS=m(ρ¨ϕ+2˙ρ˙ϕ).

Therefore the radial and transverse components of the acceleration are (¨ρρ˙ϕ2) and (ρ¨ϕ+2˙ρ˙ϕ) respectively.

We can do exactly the same thing to find the acceleration components in three-dimensional spherical coordinates. Let R, S and F be the radial, meridional and azimuthal (i.e. in direction of increasing r, θ and ϕ) components of a force on a particle.

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  • If r increases by δr, the work on the particle done is Rδr.
  • If θ increases by δθ, the work done on the particle is Srδθ.
  • If ϕ increases by δϕ, the work done on the particle is Frsinθδϕ.

Therefore Pr=R,Pθ=Sr and Pϕ=Frsinθ.

Start:

T=12m(˙r2+r2˙θ2+r2sin2θ˙ϕ2)

If you now apply Equation 13.4.12 in turn to the coordinates r,θ and ϕ, you obtain

Pr=m(¨rr˙θ2rsin2θ˙ϕ2),

Pθ=m(r2¨θ+2r˙r˙θr2sinθcosθ˙ϕ2)

and

Pϕ=m(r2sin2θ¨ϕ+2r2˙θ˙ϕsinθcosθ+2r˙r˙ϕsin2θ).

Therefore

R=m(¨rr˙θ2rsin2θ˙ϕ2),

S=m(r¨θ+2˙r˙θrsinθcosθ˙ϕ2)

and

F=m(rsinθ¨ϕ+2r˙θ˙ϕcosθ+2˙r˙ϕsinθ).

Thus the acceleration components are

  • Radial: ¨rr˙θ2rsin2θ˙ϕ2
  • Meridional: r¨θ+2˙r˙θrsinθcosθ˙ϕ2
  • Azimuthal: 2˙r˙ϕsinθ+2r˙θ˙ϕcosθ+rsinθ¨ϕ.

Be sure to check the dimensions. Since dot has dimension T-1, and these expressions must have the dimensions of acceleration, there must be an r and two dots in each term.


This page titled 13.5: Acceleration Components is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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