Question

$AB$ is a light rigid rod rotating with a constant angular speed about a vertical axis passing through $A$. 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 (wr.t. the rod) and the spring is unstrecthed. Select the correct option(s)

• A. The maximum velocity attained by the bead wr.t the rod is given by $V_{max}=\sqrt{\frac{m{\omega }^4{t}^2}{k-m{\omega }^2}}$
• B. The maximum velocity attained by the bead wr.t the rod is given by $V_{max}=\sqrt{\left(\frac{m{\omega }^4+K}{m{\omega }^2-K}\right){\omega }^2{\ell }^2}$
• C. The maximum extension in the spring is given by $X_{max}=\frac{2m{\omega }^2\ell }{K-m{\omega }^2}$
• D. The maximum value of contact force between the bead and the rod is greater than mg

Answer 1

Answer: (A), (C)

The maximum extension in the spring is given by $X_{max}=\frac{2m{\omega }^2\ell }{K-m{\omega }^2}$

Velocity will be maxium at equilibrium

$\Rightarrow m{\omega }^2(\ell +x)=Kx$
$\Rightarrow x=\frac{m{\omega }^2\ell }{K-m{\omega }^2}$
$$\frac{1}{2}m{V}_{max}^2={\int }_0^{x}m{\omega }^2(\ell +x)dx-\frac{1}{2}K{x}^2$$
$\begin{array}{cc}\Rightarrow & V_{max}^2=\frac{2m{\omega }^2\ell x+m{\omega }^2{x}^2-K{x}^2}{m}\\ & V_{max}^2=\frac{(m{\omega }^2\ell +m{\omega }^2(\ell +x)-Kx)x}{m}\\ \Rightarrow V_{max}^2& ={\omega }^2\ell =\frac{m{\omega }^4{\ell }^2}{m{\omega }^2-K}\end{array}$

$$V_{max}=\sqrt{\frac{m{\omega }^4{\ell }^2}{K-m{\omega }^2}}$$
For maximum extension
$${\int }_0^{x}m{\omega }^2(\ell +x)dx-\frac{1}{2}K{x}_{max}^2=0\Rightarrow x_{max}=\frac{2m{\omega }^2\ell }{K-m{\omega }^2}$$