Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionDynamicsEasy2 minPYQ_2023

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A body of mass $200 g$ is tied to a spring of spring constant $12.5 N m^{- 1}$ , while the other end of spring is fixed at point $O$ . If the body moves about $O$ in a circular path on a smooth horizontal surface with constant angular speed $5 rad s^{- 1}$ , then the ratio of extension in the spring to its natural length will be :

Options:

Answer:

Solution:

Let the extension in spring be $x$ and the natural length of spring be $l$ .

Here the centrifugal force will stretch the spring.

So, centrifugal force $=$ spring force

$k x = m (l + x) ω^{2} \Rightarrow 12.5 x = \frac{1}{5} (l + x) 25 \Rightarrow 1.5 x = l \Rightarrow \frac{x}{l} = \frac{1}{1.5} = \frac{2}{3}$

$\Rightarrow x : l = 2 : 3$

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Dynamics

⏱ 2mℹ️ Source: PYQ_2023

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