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Asked by Prashant DIGHE | 29 Sep, 2019, 09:54: PM

Expert Answer

Let a chain of length l is kept on a sphere so that one end of chain is fixed at top of sphere and the other end is hanging.

Let us consider a small length dl of chain that makes an angle θ with the reference horizontal line passing through centre of sphere.

Let angle subtended by this small length dl be dθ.

If m is masss of length and mass is distributed uniformly, we have linear mass density of chain = (m/l )

mass of length dl = ( R dθ )× (m/l)

Potential energy = ( R dθ )× (m/l) ×g ×Rsinθ = [ ( m g R² )/ l ] sinθ dθ

Potential energy of whole chain =

begin mathsize 14px style fraction numerator m space g space R squared over denominator l end fraction integral subscript straight pi divided by 2 space minus space straight theta subscript straight l end subscript superscript straight pi divided by 2 end superscript sin theta space d theta space equals space fraction numerator m space g space R squared over denominator l end fraction sin open parentheses l over R close parentheses end style

.....................(1)

where θ_l = (l/R) is the angle subtended by chain at centre of sphere.

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Kinetic energy of the chain when it slides by an angle θ is obtained from the difference of potential energy.

If we find the potential energies when the top end of chain was at top of sphere and when it slides an angle θ,

then this difference of potential energies gives kinetic energy at position θ.

potential energy in new position is obtained by changing the integration limits in eqn.(1)

Potential energy at new position =

begin mathsize 14px style fraction numerator m space g space R squared over denominator l end fraction integral subscript straight pi divided by 2 space minus space left parenthesis straight theta subscript straight l plus straight theta right parenthesis end subscript superscript straight pi divided by 2 space minus space theta end superscript sin theta space d theta space equals space fraction numerator m space g space R squared over denominator l end fraction space open square brackets sin open parentheses theta subscript l plus theta close parentheses space minus space sin theta space close square brackets space space.............. left parenthesis 2 right parenthesis end style

Kinetic energy is given by

begin mathsize 14px style 1 half m v squared space equals space fraction numerator m space g space R squared over denominator l end fraction open square brackets sin space theta subscript l space minus space sin open parentheses theta subscript l plus theta close parentheses space plus space sin theta close square brackets end style

...........................(3)

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For tangential acceleration, we get the eqn. for velocity v from eqn.(3) and differentiate.

we get, dv/dt

begin mathsize 14px style equals space fraction numerator space g space R squared over denominator l space v end fraction open square brackets cos space theta space minus space cos open parentheses theta space plus space theta subscript l close parentheses close square brackets fraction numerator d theta over denominator d t end fraction end style

Answered by Thiyagarajan K | 30 Sep, 2019, 04:44: PM

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