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Verify equation (18.3) i.e.ld2θdl2+2»åθdl+gV2θ=0

Short Answer

Expert verified

The resultant answer ld2θdl2+2»åθdl+gV2θ=0is verified.

Step by step solution

01

Concept of Equation of motion:

The equation of motion:

ddt(ml2θ)=³¾²µ±ô²õ¾±²Ôθ

02

Consider jn(x)=∑r=0∞(-1)ΓΓ(r+1)Γ(r+1+n)(x2)2Γ+n and simplify it:

Considering the equation of motion:

ddtml2θ=mgl²õ¾±²Ôθ

For the small value of θ:

sin=θ

Hence,

ml2d2θdl2+2lmdldt×»åθdt+³¾²µ±ôθ=0 ….. (1)

Consider the following formula.

l=l0+vt

Take a derivation with respect to t .

role="math" localid="1659272604614" dldt=vdl=vdt

Then,

»åθdt=»åθdl×dldt=v»åθdt.

And,

d2θdt2=V2d2θdl2.

03

Put the values into equation (1):

ml2v2d2θdl2+2lmv×v»åθdt+³¾²µ±ôθv2v2=0mlv2ld2θdl2+2»åθdt+gv2θ=0ld2θdl2+2»åθdt+gv2θ=0

Hence, ld2θdl2+2»åθdl+gV2θ=0 is verified.

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