Physics - Rotation Question with Solution | TestHub

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$\text{Δθ}$at the centre





$\text{ΔN sin θ = Δmg}$...... (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$\text{Δ θ }$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

$\lambda =\frac{m}{2\pi }$

∴$\Delta m=\lambda \Delta \theta =\frac{m}{2\pi }\Delta \theta =\frac{m\Delta \theta }{2\pi }$

∴$\text{cot}\theta =\frac{\text{T}\Delta \theta -\left(\frac{m\Delta \theta }{2\pi }\right)R{\omega }^2}{\left(\frac{m}{2\pi }\Delta \theta \right)g}$

∴$T=\left(R{\omega }^2+\text{g}\text{cot}\theta \right)\frac{m}{2\pi }$