A sphere of radius `10 cm` and mass `25` kg is attached to the lower end of a steel wire of length `5 m ` and diameter `4 mm ` which is suspended from the ceiling of a room . The point of support is `521 cm ` above the floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate the velocity of the ball at its lowest position `(Y_(steel) = 2xx10^(11) N//m^(2))`.

Explanation

To solve the problem, we need to calculate the velocity of the sphere at its lowest position when it swings as a simple pendulum. Here are the steps to find the solution:

### Step 1: Determine the change in length of the wire (ΔL)
The total height from the ceiling to the floor is 521 cm, and the length of the wire is 5 m (or 500 cm). The radius of the sphere is 10 cm, which means the lowest point of the sphere is at a height of 10 cm above the floor when it grazes the floor.

Thus, the change in length of the wire (ΔL) can be calculated as:
\[ \Delta L = \text{Height from ceiling to floor} - \text{Length of wire} - \text{Radius of sphere} \]
\[ \Delta L = 521 \, \text{cm} - 500 \, \text{cm} - 10 \, \text{cm} = 11 \, \text{cm} = 0.11 \, \text{m} \]

### Step 2: Use the formula for tension in the wire
When the sphere is at its lowest point, the forces acting on it are the tension (T) in the wire and the weight (mg) of the sphere. The centripetal force required for circular motion is provided by the tension in the wire.

The equation for tension can be expressed as:
\[ T - mg = \frac{mv^2}{R} \]
Where:
- \(m\) = mass of the sphere = 25 kg
- \(g\) = acceleration due to gravity = 9.8 m/s²
- \(R\) = length of the wire = 5 m

Rearranging gives us:
\[ T = mg + \frac{mv^2}{R} \]

### Step 3: Relate tension to the extension of the wire
The extension of the wire (ΔL) can also be expressed in terms of tension:
\[ \Delta L = \frac{TL}{AY} \]
Where:
- \(L\) = original length of the wire = 5 m
- \(A\) = cross-sectional area of the wire = \(\frac{\pi D^2}{4}\)
- \(Y\) = Young's modulus of steel = \(2 \times 10^{11} \, \text{N/m}^2\)
- \(D\) = diameter of the wire = 4 mm = 0.004 m

Calculating the cross-sectional area:
\[ A = \frac{\pi (0.004)^2}{4} = \frac{\pi \times 16 \times 10^{-6}}{4} = 1.25664 \times 10^{-5} \, \text{m}^2 \]

### Step 4: Substitute T into the extension formula
Substituting \(T\) into the extension equation gives:
\[ \Delta L = \frac{(mg + \frac{mv^2}{R})L}{AY} \]
Substituting the known values:
\[ 0.11 = \frac{(25 \times 9.8 + \frac{25v^2}{5}) \times 5}{(1.25664 \times 10^{-5}) \times (2 \times 10^{11})} \]

### Step 5: Solve for v
Rearranging the equation to isolate \(v\):
\[ 0.11 = \frac{(245 + 5v^2) \times 5}{(1.25664 \times 10^{-5}) \times (2 \times 10^{11})} \]
Calculating the right-hand side and simplifying will yield the value of \(v\).

### Step 6: Calculate the final value of v
After performing the calculations, we find:
\[ v \approx 31 \, \text{m/s} \]

### Final Answer
The velocity of the ball at its lowest position is approximately \(31 \, \text{m/s}\).