JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 10)
A particle of mass m is fixed to one end of a
light spring having force constant 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 :
$${{m\ell {\omega ^2}} \over {k - m{\omega ^2}}}$$
$${{m\ell {\omega ^2}} \over {k - m{\omega}}}$$
$${{m\ell {\omega ^2}} \over {k + m{\omega ^2}}}$$
$${{m\ell {\omega ^2}} \over {k + m{\omega}}}$$
Explanation
_8th_January_Morning_Slot_en_10_1.png)
At elongated position (x),
Fradial = mr$${\omega ^2}$$
$$ \therefore $$ kx = m$$\left( {{l} + x} \right)$$$${\omega ^2}$$
$$ \Rightarrow $$ x = $${{m\ell {\omega ^2}} \over {k - m{\omega ^2}}}$$
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