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4-1.Newton's Laws of Motion
normal
A particle of mass $\mathrm{m}$ is fixed to one end of a light spring having force constant $\mathrm{k}$ and unstretched length $\ell .$ The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is
A$\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}+\mathrm{m} \omega^{2}}$
B$\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}-\mathrm{m} \omega^{2}}$
C$\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}-\mathrm{m} \omega}$
D$\frac{\mathrm{m} \ell \omega^{2}}{\mathrm{k}+\mathrm{m} \omega}$
Solution
$k\,x = m\,{\omega ^2}r$$k\,x = m\,{\omega ^2}(l + x)$
$k\,x = m{\omega ^2}l + m\,{\omega ^2}x$
$x(k – m\,{\omega ^2}) = m\,{\omega ^2}l$
$\therefore \,\,x = \frac{{m\,{\omega ^2}l}}{{k – m\,{\omega ^2}}}$
Standard 11
Physics
