Question

A chain of mass $m$ forming a circle of radius $R$ is slipped on a smooth round cone with half-angle $\theta$ . Find the tension in the chain if it rotates with a constant angular velocity $\omega$ about a vertical axis coinciding with the symmetry axis of the cone.

• A. $\left(R{\omega }^2+g\cot \theta \right)\frac{m}{2\pi }$
• B. $\left(R{\omega }^2-\mathrm{g}\cot \theta \right)\frac{m}{2\pi }$
• C. $\left(R{\omega }^2+\mathrm{g}\cot \theta \right)\frac{m}{\pi }$
• D. $(R\omega -\mathrm{g}\cot \theta )\frac{m}{2\pi }$

Answer 1

Answer: (A)

Every element of chain moves on a circular path. So, every element of the chain experiences centripetal force. We consider a small element $\Delta \text{m}$ on the chain making angle $\Delta \theta$ at the centre


$\Delta N\sin \theta =\Delta mg$ ...... (i)
Net force towards centre is
$2\text{T}\text{sin}\frac{\Delta \theta }{2}-\Delta \text{N}\text{cos}\theta =\Delta \text{mR}{\omega }^2$
$\Delta N\text{cos}\theta =2T\text{sin}\frac{\Delta \theta }{2}-\Delta \text{mR}{\omega }^2$ .......... (ii)
From eqn.(i) and (ii) we get
$\text{cos}\theta =\frac{2\text{T}\text{sin}\frac{\Delta \theta }{2}-\Delta \text{mR}{\omega }^2}{\Delta \text{mg}}$
But $\Delta \theta$ is very small.
∴ $\text{sin}\frac{\Delta \theta }{2}=\frac{\Delta \theta }{2}$
∵ $\text{cot}\theta =\frac{\text{T}\Delta \theta -\Delta \text{mR}{\omega }^2}{\Delta \text{mg}}$
Here Δm makes an angle Δθ at the centre. The total mass of chain is m.
∴ The mass of chain per unit angle is
$\lambda =\frac{m}{2\pi }$
∴ $\Delta m=\lambda \Delta \theta =\frac{m}{2\pi }\Delta \theta =\frac{m\Delta \theta }{2\pi }$
∴ $\text{cot}\theta =\frac{\text{T}\Delta \theta -\left(\frac{m\Delta \theta }{2\pi }\right)R{\left(\omega \right)}^2}{\left(\frac{m}{2\pi }\Delta \theta \right)g}$
$\therefore T=\left(R{\omega }^2+g\cot \theta \right)\frac{m}{2\pi }$