Question

AB is a light rigid rod, which is rotating 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 (w.r.t. the rod) and the spring is unstretched.

Select correct options:

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

Answer 1

Answer: (A), (C)

The maximum velocity attained by the bead w.r.t the rod is given by $V_{max}=\sqrt{\frac{m{\omega }^4{l}^2}{K-m{\omega }^2}}$
The maximum extension in the spring is given by $X_{max}=\frac{2m{\omega }^{2}l}{K-m{\omega }^2}$

$m{\omega }^2(l+x)=Kx$
$X=\frac{m{\omega }^{2}l}{K-m{\omega }^2}$
Velocity will be maximum at equilibrium position.
$\frac{1}{2}m{V}_{max}^2={\int }_0^{x}m{\omega }^2(l+x)dx-\frac{1}{2}K{x}^2$
$V_{max}^2=\frac{2m{\omega }^{2}lx+m{\omega }^2{x}^2-K{x}^2}{m}$

$V_{max}^2=\frac{(m{\omega }^{2}l+m{\omega }^2(l+x)-Kx)x}{m}$

$V_{max}^2={\omega }^{2}lx=\frac{m{\omega }^4{l}^2}{m{\omega }^2-K}$

$V_{max}=\sqrt{\frac{m{\omega }^4{l}^2}{K-m{\omega }^2}}$

For maximum extension

${\int }_0^{x_{max}}m{\omega }^2(l+x)dx-\frac{1}{2}K{x}_{max}^2=0$

$X_{max}=\frac{2m{\omega }^{2}l}{K-m{\omega }^2}$