Question 127: Work, Energy and Power & Circular Motion

Question

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 :

Accepted Answer

At elongated position (x),
F
radial
= mr$${\omega ^2}$$
$$ 	herefore $$ kx = m$$\left( {{l} + x} ight)$$$${\omega ^2}$$
$$ \Rightarrow $$ x = $${{m\ell {\omega ^2}} \over {k - m{\omega ^2}}}$$

Answer

$${{m\ell {\omega ^2}} \over {k - m{\omega ^2}}}$$

Answer

$${{m\ell {\omega ^2}} \over {k - m{\omega}}}$$

Answer

$${{m\ell {\omega ^2}} \over {k + m{\omega ^2}}}$$

Answer

$${{m\ell {\omega ^2}} \over {k + m{\omega}}}$$

Work, Energy and Power & Circular Motion

Choose the correct answer: