One end of a massless spring of spring constant k and natural length… | JEE Main 2022 (27 Jun, Shift 2)

Q36JEE Main 2022 · 27 JUN · SHIFT 2Medium

Rotational Kinematics & Dynamics

One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing through fixed end, then the elongation of the spring will be:

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Answer: (c)

Spring force gives the centripetal force; radius of circle $= l_0 + x$ .
$kx = m\omega^2 (l_0 + x)$
$x(k - m\omega^2) = m\omega^2 l_0 \Rightarrow x = \dfrac{m\omega^2 l_0}{k - m\omega^2}$
So the answer is (c).

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