Physics - Mechanical Properties of Solids Question with Solution | TestHub

PhysicsMechanical Properties of SolidsElasticityMedium3 minCT48

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Passage / Comprehension

a ring of mass m and radius R rolls with constant speed v on a horizontal surface. Its material has Young modulus y and the wire cross-sectional area is A.

Tensile force applied by one half of the ring on the other half, due to its motion is :-

Options:

Answer:

Solution:

In the centre-of-mass frame, the ring rotates about its centre with $ω = \frac{v}{R}$ .

The CM of the half-ring (mass $\frac{m}{2}$ ) is at distance $\frac{2 R}{π}$ from the centre, so it needs centripetal force $F = \frac{m}{2} ω^{2} \frac{2 R}{π} = \frac{m v ^{2}}{π R}$ .

The two cut ends supply this through tension: $2 T = \frac{m v ^{2}}{π R} \Rightarrow T = \frac{m v ^{2}}{2 π R}$ - option (D).

Stream:JEESubject:PhysicsTopic:Mechanical Properties of SolidsSubtopic:Elasticity

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