A particle moves in a circular orbit of radius ' r ' under a central attractive force, F = − r k , where k is a constant. The periodic time of tis motion is

mr ω 2 = r k ω 2 = mr k ω 2 ∝ r 2 1 ω ∝ r 1 ∴ T ∝ r

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Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionMiscellaneousEasy2 minPYQ_2021

PhysicsEasysingle choice

A particle moves in a circular orbit of radius ' $r$ ' under a central attractive force, $F = - \frac{k}{r}$ , where $k$ is a constant. The periodic time of tis motion is

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Solution:

$mr ω^{2} = \frac{k}{r}$ $ω^{2} = \frac{k}{mr}$ $ω^{2} \propto \frac{1}{r ^{2}}$ $ω \propto \frac{1}{r}$ $∴ T \propto r$

Stream:MHT_CETSubject:PhysicsTopic:Circular MotionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2021

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