A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1m, is whirled in a vertical circle with an angular velocity of `2 rev.//s` at the bottom of the circle. The cross-sectional area of the wire is `0.065 cm^(2)` . Calculate the elongation of the wire when the mass is at the lowest point of its path `Y_(steel) = 2 xx 10^(11) Nm^(-2)`.

Options

A. 0.52 mm
B. 1.87 mm
C. 2.52 mm
D. 3.52 mm

Explanation

To solve the problem of calculating the elongation of the steel wire when a 14.5 kg mass is whirled in a vertical circle, we will follow these steps:

### Step 1: Calculate the Forces Acting on the Wire
At the lowest point of the circular path, the total force acting on the wire is the sum of the gravitational force (weight of the mass) and the centripetal force required to keep the mass moving in a circle.

1. **Weight of the mass (mg)**:
\[ m = 14.5 \, \text{kg}, \quad g = 9.8 \, \text{m/s}^2 \]
\[ F_g = mg = 14.5 \times 9.8 = 142.1 \, \text{N} \]

2. **Centripetal force (mω²r)**:
First, we need to calculate the angular velocity (ω) in radians per second.
\[ \text{Angular velocity} \, \nu = 2 \, \text{rev/s} = 2 \times 2\pi \, \text{rad/s} = 4\pi \, \text{rad/s} \]
Now, we can calculate the centripetal force:
\[ r = 1 \, \text{m} \quad (\text{length of the wire}) \]
\[ F_c = m \cdot r \cdot \omega^2 = 14.5 \cdot 1 \cdot (4\pi)^2 \]
\[ F_c = 14.5 \cdot 16\pi^2 \approx 14.5 \cdot 157.91 \approx 2293.7 \, \text{N} \]

3. **Total force (F)**:
\[ F = F_g + F_c = 142.1 + 2293.7 \approx 2435.8 \, \text{N} \]

### Step 2: Calculate the Elongation of the Wire
Using Young's modulus (Y) to find the elongation (ΔL) of the wire:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
Where:
- \( F \) = total force = 2435.8 N
- \( L \) = original length of the wire = 1 m
- \( A \) = cross-sectional area = 0.065 \, \text{cm}^2 = 0.065 \times 10^{-4} \, \text{m}^2 = 6.5 \times 10^{-6} \, \text{m}^2
- \( Y \) = Young's modulus for steel = \( 2 \times 10^{11} \, \text{N/m}^2 \)

Rearranging the formula to solve for ΔL:
\[ \Delta L = \frac{F \cdot L}{Y \cdot A} \]
Substituting the values:
\[ \Delta L = \frac{2435.8 \cdot 1}{2 \times 10^{11} \cdot 6.5 \times 10^{-6}} \]
Calculating:
\[ \Delta L = \frac{2435.8}{1.3 \times 10^{6}} \approx 1.87 \times 10^{-3} \, \text{m} \]
Converting to millimeters:
\[ \Delta L \approx 1.87 \, \text{mm} \]

### Final Answer
The elongation of the wire when the mass is at the lowest point of its path is approximately **1.87 mm**.

---