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 `theta` with the vertical and is released. Find the maximum value of `theta` so that the sphere does not rub the floor. young's modulus of the metal of the wire is `2.0xx10^11 Nm^-2`. Make appropriate approximation.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Problem
We have a sphere of mass \( m = 20 \, \text{kg} \) suspended by a wire of length \( L = 4 \, \text{m} \) and diameter \( d = 1 \, \text{mm} \). There is a gap of \( 2 \, \text{mm} \) between the sphere and the floor when in equilibrium. We need to find the maximum angle \( \theta \) such that the sphere does not touch the floor when it is pushed aside and released.

### Step 2: Calculate the Radius of the Wire
The radius \( R \) of the wire can be calculated from the diameter:
\[ R = \frac{d}{2} = \frac{1 \, \text{mm}}{2} = 0.5 \, \text{mm} = 5 \times 10^{-4} \, \text{m} \]

### Step 3: Calculate the Force Acting on the Wire
The force acting on the wire due to the weight of the sphere is given by:
\[ F = mg = 20 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 196.2 \, \text{N} \]

### Step 4: Calculate the Area of the Wire
The cross-sectional area \( A \) of the wire is given by:
\[ A = \pi R^2 = \pi (5 \times 10^{-4} \, \text{m})^2 = \pi \times 25 \times 10^{-8} \, \text{m}^2 \approx 7.85 \times 10^{-8} \, \text{m}^2 \]

### Step 5: Calculate the Young's Modulus and Elongation
Using Young's modulus \( Y \) and the formula for elongation \( \Delta L \):
\[ Y = \frac{F L}{A \Delta L} \]
Rearranging gives:
\[ \Delta L = \frac{F L}{A Y} \]
Substituting the known values:
- \( F = 196.2 \, \text{N} \)
- \( L = 4 \, \text{m} \)
- \( A \approx 7.85 \times 10^{-8} \, \text{m}^2 \)
- \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)

Calculating \( \Delta L \):
\[ \Delta L = \frac{196.2 \times 4}{7.85 \times 10^{-8} \times 2.0 \times 10^{11}} \approx \frac{784.8}{1.57 \times 10^4} \approx 0.050 \, \text{m} = 50 \, \text{mm} \]

### Step 6: Maximum Elongation Condition
The maximum elongation allowed is \( 2 \, \text{mm} \) (the gap between the sphere and the floor). Therefore, we need to find the condition when the elongation is equal to \( 2 \, \text{mm} \) or \( 0.002 \, \text{m} \).

### Step 7: Relate the Angle \( \theta \) to the Height
When the sphere is displaced to an angle \( \theta \), the vertical height \( h \) from the top of the wire to the center of the sphere is given by:
\[ h = L - L \cos \theta = L(1 - \cos \theta) \]
The elongation \( \Delta L \) can be expressed as:
\[ \Delta L = h - 0.002 \, \text{m} \]
Setting \( \Delta L = 0.002 \, \text{m} \):
\[ L(1 - \cos \theta) - 0.002 = 0.002 \]
\[ L(1 - \cos \theta) = 0.004 \]
Substituting \( L = 4 \, \text{m} \):
\[ 4(1 - \cos \theta) = 0.004 \]
\[ 1 - \cos \theta = 0.001 \]
\[ \cos \theta = 0.999 \]

### Step 8: Calculate \( \theta \)
Using the cosine inverse function:
\[ \theta = \cos^{-1}(0.999) \approx 0.0573 \, \text{radians} \approx 3.29^\circ \]

### Final Answer
The maximum value of \( \theta \) so that the sphere does not rub the floor is approximately \( 3.29^\circ \).

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