Laws of Motion

JEE Main level, single correct, medium

116 of 210

Question 116 of 210

One end of a massless spring of spring constant kk and natural length l0{l}_{0} is fixed while the other end is connected to a small object of mass mm 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 trough fixed end, then the elongation of the spring will be

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Answer: (C) mω2l0k−mω2\dfrac{m{\omega }^{2}{l}_{0}}{k-m{\omega }^{2}}

The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude v2r\dfrac{{v}^{2}}{r}. The horizontal force on the particle is due to the spring and is given by kxkx, where xx is the elongation and kk is the spring constant.
kx=mv2r=mω2r=mω2(l0+x)kx=\dfrac{m{v}^{2}}{r}=m{\omega }^{2}r=m{\omega }^{2}\left({l}_{0}+x\right)
⇒(k−mω2)x=mω2l0\Rightarrow \left(k-m{\omega }^{2}\right)x=m{\omega }^{2}{l}_{0}
⇒x=mω2l0k−mω2\Rightarrow x=\dfrac{m{\omega }^{2}{l}_{0}}{k-m{\omega }^{2}}

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