NEET Class neet Answered

how to solve this by taking components

Asked by Prashant DIGHE | 13 Apr, 2020, 09:50: PM

Expert Answer

Let us consider a small mass element of length dl in the ring .

If mass is distributed unifromly, let us consider linear mass density , i.e. mass per unit length ρ and this length dl has mass ρdl.

Gravitational force F experienced by element of length dl in the ring due to mass m is given by,

F = G ( ρdl × m )/ d²

This force can be resolved into two components, one component Fsinθ in the plane of ring and

other component Fcosθ perpendicular to the plane of ring.

If we integrate the force over the full circumference of ring, Fsinθ component will be cancelled out.

Hence force on full ring =

begin mathsize 14px style integral G space cos theta space fraction numerator rho cross times m over denominator d squared end fraction space d l space space equals space G fraction numerator rho cross times m over denominator d squared end fraction space left parenthesis space 2 pi R right parenthesis space cos theta space equals space G fraction numerator M cross times m over denominator d squared end fraction cos theta end style

Where M = (2πR)ρ is mass of ring. we have, d = (5/3)R , cosθ = 4/5

Force on ring is given by,

begin mathsize 14px style F space equals space 36 over 125 cross times G cross times fraction numerator M cross times m over denominator R squared end fraction end style

Answered by Thiyagarajan K | 13 Apr, 2020, 11:36: PM

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