Question

A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, 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 θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young modulus of the metal of the wire is 2.0 × 10¹¹ N m⁻². Make appropriate approximations.

Answer 1

Given:
Mass of sphere (m) = 20 kg
Length of metal wire (L) = 4 m
Diameter of wire (d = 2r) = 1 mm
⇒ r = 5 × 10⁻⁴ m
Young's modulus of the metal wire = 2.0 × 10¹¹ N m⁻²
Tension in the wire in equilibrium = T
T= mg
When it is moved at an angle θ and released, let the tension at the lowest point be T'​.
$\Rightarrow T\text{'}=mg+\frac{m{v}^2}{r}$
The change in tension is due to the centrifugal force.
∴​ $∆T=T\text{'}-T$
$∆T=\frac{m{v}^2}{r}...\left(\mathrm{i}\right)$

Now, using work energy principle:
$$\begin{gathered} \frac{1}{2}m{v}^2-0=mgr\left(1-\cos \theta \right) \\ \Rightarrow v^2=2gr\left(1-\cos \theta \right)...\left(2\right) \end{gathered}$$

Applying the value of v² in (i):

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

$$\begin{gathered} \mathrm{Now},F=∆T \\ \mathrm{Also},F=\frac{YA∆L}{L} \\ \Rightarrow \frac{YA∆L}{L}=2mg\left(1-\cos \theta \right) \\ \Rightarrow \cos \theta =1-\frac{YA∆L}{L\left(2mg\right)} \\ \Rightarrow cos\theta =1-\left[\frac{2\times {10}^{11}\times 4\times 3.14\times {\left(5\right)}^2\times {10}^{-8}\times 2\times {10}^{-3}}{4\times 2\times 20\times 10}\right] \\ \Rightarrow \cos \theta =0.80 \\ \mathrm{Or},\theta =36.4° \end{gathered}$$

Hence, the required maximum value of θ is 35.4˚.