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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassImpulse momentum theoremMedium2 minQB

PhysicsMediumsingle choice

A particle of mass $m = 1 kg$ moves in a circle of radius $R = 2 m$ with uniform speed $v = 3 π m / s$ . The magnitude of impulse given by centripetal force to the particle in one second is :-

Options:

Answer:

Solution:

$ω = \frac{v}{R} = \frac{3 π}{2} rad/s$ , so in 1 s the particle turns through 270°. The impulse is the momentum change:

$∣Δ p ∣ = 2 m v sin 13 5^{\circ} = 2 \times 1 \times 3 π \times \frac{2}{2} = 32 π Ns$

Ans. (D)

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Impulse momentum theorem

⏱ 2mℹ️ Source: QB

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