Question

A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4m and diameter 1 mm. When in equilbrium, there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle $\theta$ with the vertical and is released. Find the maximum value of $\theta$ so that the sphere does not rub the floor. Young modulus of the metal of the wire is $2.0\times {10}^{11}N{m}^{-2}.$ Make appropriate approximations.

Answer 1

m = 20 kg,

L = 4m ,

2r = 1 mm , r = $5\times {10}^{-4}m$.

At equilibrium , T = mg

When it moves at an angle $\theta$ and released, the tension 'T' at lowest point is

$\Rightarrow T=mg+\frac{m{v}^2}{r}$

The change in tension is due to centrifugal force

$\Delta T=\frac{m{v}^2}{r}$ ...(1)

$\Rightarrow$ Again , by work energy principle

$\frac{1}{2}m{v}^2-0=mgr(1-cos\theta )$

$\Rightarrow v^2=2gr(1-cos\theta )$ ....(2)

So, $\Delta T=\frac{m\left[2gr\right(1-cos\theta \left)\right]}{r}$

= 2 mg $(1-cos\theta )$

$\Rightarrow F=\Delta T$

$\Rightarrow F=\frac{YA\Delta L}{L}$

$=2mg-2mgcos\theta$

$\Rightarrow 2mgcos\theta =2mg-\frac{YA\Delta L}{L}$

$\Rightarrow cos\theta =1-\frac{YA\Delta L}{L\left(2mg\right)}$

$\Rightarrow cos\theta =1-\left[\frac{2\times {10}^{13\times 4\times 3.14}\times (5{)}^2\times {10}^{-8}\times 2\times {10}^{-3}}{4\times 2\times 20\times 10}\right]$

$\Rightarrow cos\theta =0.80$

$\theta ={36.4}^{\circ }$