Question

A sphere of mass 20kg is suspended by a metal wire of unstretched length 4 m and diameter 1mm.When is equilibrium, there is a clear gap of 2mm between the so that the wire makes an angle $\theta$ with the vetical 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

At equilibrium $\Rightarrow T=mg$

When it moves to an angle $E$, and released, the tension the $T^{\prime }$ at lowest point is

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

The change in tension is due to centrifugal force

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

Again, by work energy theorem,

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

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

So,simplifying the above equation:

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

The force acting on the string will be the change in the tension.

$\Rightarrow F=\Delta T$

Using the relation between the stress,strain and young's modulus:

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

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

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