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JEE Main 2022 (27 Jun, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from 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 ω\omega about an axis passing through fixed end, then the elongation of the spring will be:

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Answer: (c)

  1. Spring force gives the centripetal force; radius of circle =l0+x= l_0 + x.
  2. kx=mω2(l0+x)kx = m\omega^2 (l_0 + x)
  3. x(k−mω2)=mω2l0⇒x=mω2l0k−mω2x(k - m\omega^2) = m\omega^2 l_0 \Rightarrow x = \dfrac{m\omega^2 l_0}{k - m\omega^2}
  4. So the answer is (c).
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