JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 24)
A spring mass system (mass m, spring
constant k and natural length $$l$$) rest in
equilibrium on a horizontal disc. The free end
of the spring is fixed at the centre of the disc.
If the disc together with spring mass system,
rotates about it's axis with an angular velocity
$$\omega $$, (k $$ \gg m{\omega ^2}$$) the relative change in the length
of the spring is best given by the option :
$${{m{\omega ^2}} \over {3k}}$$
$${{m{\omega ^2}} \over k}$$
$${{2m{\omega ^2}} \over k}$$
$$\sqrt {{2 \over 3}} \left( {{{m{\omega ^2}} \over k}} \right)$$
Explanation
m$${\omega ^2}$$(l0 + x) = kx
x = $${{m{I_0}{\omega ^2}} \over {k - m{\omega ^2}}}$$
For k >> m$${\omega ^2}$$
$${x \over {{I_0}}} = {{m{\omega ^2}} \over k}$$
x = $${{m{I_0}{\omega ^2}} \over {k - m{\omega ^2}}}$$
For k >> m$${\omega ^2}$$
$${x \over {{I_0}}} = {{m{\omega ^2}} \over k}$$
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