Question

A small sphere of mass $200gm$ is attached to an inextensible string of length $130cm$ whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius $50cm$. Calculate the time period of this conical pendulum and the tension in the string. $({\pi }^2=10)$

Answer 1

Balancing the forces:

$T\cos \theta =mg⟶\left(1\right)T\sin \theta =\frac{m{v}^2}{r}⟶\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$

$\tan \theta =\frac{v^2}{rg}(\sin \theta =\frac{r}{l}=\frac{50}{130}\theta ={sin}^{-1}\left(\frac{5}{13}\right)=22.62°)\Rightarrow v^2=rg\tan \theta \Rightarrow v=\sqrt{rg\tan \theta }=\sqrt{50\times 10\times 0.42\times {10}^{-2}}$

$=\sqrt{2.1}=145m/s$

Time period $T=\frac{2\pi }{\omega }=\frac{2\pi r}{v}T=\frac{2\pi r}{v}=\frac{2\pi \times 50\times {10}^{-2}}{1.45}=2.16s$

To calculate tension:

$T^2{cos}^2\theta +T^2{sin}^2\theta ={\left(mg\right)}^2+{\left(\frac{m{v}^2}{r}\right)}^2{T}^2={\left(mg\right)}^2+{\left(\frac{m{v}^2}{r}\right)}^{2}T=\sqrt{{\left(mg\right)}^2+{\left(\frac{m{v}^2}{r}\right)}^2}=\sqrt{{(0.2\times 10)}^2+{\left(\frac{0.2\times 2.10}{50\times {10}^{-2}}\right)}^2}=\sqrt{4+0.70}=\sqrt{4.7}=2.17N$