One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing through fixed end, then the elongation of the spring will be :
Solution
<p>$m{\omega ^2}({l_0} + x) = kx$</p>
<p>$\Rightarrow m{\omega ^2}{l_0} = (k - m{\omega ^2}) \times x$</p>
<p>$\Rightarrow x = {{m{\omega ^2}{l_0}} \over {(k - m{\omega ^2})}}$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
This question is part of PrepWiser's free JEE Main question bank. 158 more solved questions on Rotational Motion are available — start with the harder ones if your accuracy is >70%.