Physics - NLM Question with Solution | TestHub

PhysicsNLMChallenging application of NLMMedium2 minQB

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Passage / Comprehension

Two identical rings, each of mass m, are free to slide on a long hoop of mass M. Both rings are released from rest at the top of the hoop. Initially the hoop stands in a vertical plane on a table. Eventually the hoop loses contact with the table as shown. All surfaces are frictionless.

Shared passage diagram

If $M = 1 kg & m = 2 kg$ than, what is the angle made by line joining either ring to centre of hoop with vertical line at a time when hoop will leave the contact:-

Options:

Answer:

Solution:

At angle θ from the top, the bead gains kinetic energy:

$\frac{1}{2} m v^{2} = m g R (1 - cos θ)$

Radial force balance on each bead gives:

$N + m g cos θ = \frac{m v ^{2}}{R}$

$N = 2 m g - 3 m g cos θ$

For the hoop of mass M, ground normal N1 falls to zero at first lift.

$N_{1} + 2 N cos θ = M g$

$N_{1} = 0 \Rightarrow 2 (2 m g - 3 m g cos θ) cos θ = M g$

$6 m cos^{2} θ - 4 m cos θ + M = 0$

$m = 2 kg, M = 1 kg$

$12 cos^{2} θ - 8 cos θ + 1 = 0$

$cos θ = \frac{1}{2} or \frac{1}{6}$

Starting from the top, cosθ first reaches 1/2, so θ=60° (A).

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Challenging application of NLM

⏱ 2mℹ️ Source: QB

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