Question

$AB$ is a light rigid rod, which is rotating about a vertical axis passing through end $A$ . A spring of force constant $k$ and natural length $l$ is attached at $A$ and its other end is attached to a small bead of mass $m$ . The bead can slide without friction on the rod. At the initial moment, the bead is at rest (with respect to the rod) and the spring is unstretched. The maximum speed of the bead with respect to the rod during the course of motion is

• A. $V_{\text{max}}=\sqrt{\frac{m{\omega }^2{l}^2}{k-m{\omega }^2}}$
• B. $V_{\text{max}}=\sqrt{\frac{m{\omega }^4{l}^2}{k-m{\omega }^2}}$
• C. $V_{\text{max}}=\sqrt{\frac{m{\omega }^4{l}^2}{m{\omega }^2-k}}$
• D. $V_{\text{max}}=\sqrt{\frac{m{\omega }^2{l}^2}{m{\omega }^2-k}}$

Answer 1

Answer: (B)

Velocity will be maximum at the equilibrium position
$\Rightarrow kx=m{\omega }^2(l+x)$
$\Rightarrow x=\frac{m{\omega }^{2}l}{k-m{\omega }^2}$
Now, using work-energy theorem

$\Delta KE=$ Work done by all the forces
$\frac{1}{2}m{V}_{\text{max}}^2=\underset{0}{\overset{x}{\int }}m{\left(\omega \right)}^2\left(l+x\right)dx-\frac{1}{2}k{x}^2$
$\Rightarrow V_{\text{max}}^2=\frac{2m{\omega }^{2}lx+m{\omega }^2{x}^2-k{x}^2}{m}$
$V_{\max }^2=\frac{\left(m{\omega }^{2}l+m{\omega }^2(l+x)-kx\right)x}{m}$
$=\frac{m{\omega }^{2}lx}{m}$
$\Rightarrow V_{\text{max}}^2={\omega }^{2}lx=\frac{m{\omega }^4{l}^2}{k-m{\omega }^2}$
$V_{\text{max}}=\sqrt{\frac{m{\omega }^4{l}^2}{k-m{\omega }^2}}$