A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4m and diameter 1mm. When in equilibrium, there is a clear gap of 2mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle α with the vertical and is released. Find the maximum value of α so that the sphere does not rub the floor. Young modulus of metal wire is 2×10¹¹ N m^-2 Make appropriate approximations.

To determine the maximum angle α that the sphere can be pushed aside without touching the floor, we need to analyze the forces acting on the sphere and the properties of the wire. The key here is to understand how the wire stretches when the sphere is displaced and how this affects the height of the sphere above the floor.

Understanding the Setup

We have a sphere with a mass of 20 kg suspended by a wire. The wire has an unstretched length of 4 m and a diameter of 1 mm. When the sphere is at rest, it hangs 2 mm above the floor. This means that the effective length of the wire when the sphere is in equilibrium is 4 m - 0.002 m = 3.998 m.

Calculating the Stretch of the Wire

When the sphere is pushed aside and released, the wire will stretch due to the tension created by the weight of the sphere. The tension in the wire can be calculated using the formula:

• Tension (T) = Weight of the sphere = mg

Here, m is the mass of the sphere (20 kg) and g is the acceleration due to gravity (approximately 9.81 m/s²). Thus, the tension is:

T = 20 kg × 9.81 m/s² = 196.2 N.

Using Young's Modulus

Young's modulus (E) relates the stress and strain in a material. The formula is given by:

• E = (T/A) / (ΔL/L₀)

Where:

• T = tension in the wire (196.2 N)
• A = cross-sectional area of the wire
• ΔL = change in length of the wire
• L₀ = original length of the wire (4 m)

The cross-sectional area (A) of the wire can be calculated using the diameter:

A = π(d/2)² = π(0.001 m / 2)² = π(0.0005 m)² ≈ 7.85 × 10⁻⁷ m².

Calculating the Change in Length

Rearranging the Young's modulus equation to find ΔL gives us:

ΔL = (T × L₀) / (E × A).

Substituting the values:

• ΔL = (196.2 N × 4 m) / (2 × 10¹¹ N/m² × 7.85 × 10⁻⁷ m²)

Calculating this yields:

ΔL ≈ 0.0005 m or 0.5 mm.

Determining the Maximum Angle α

When the sphere is displaced to an angle α, the vertical height of the sphere above the floor can be expressed as:

Height = L₀ - ΔL - (L₀ × cos(α)).

For the sphere to not touch the floor, this height must be greater than 2 mm:

L₀ - ΔL - (L₀ × cos(α)) > 0.002 m.

Substituting the known values:

4 m - 0.0005 m - (4 m × cos(α)) > 0.002 m.

Solving for cos(α):

3.9995 m - 4 m × cos(α) > 0.002 m

4 m × cos(α) < 3.9975 m

cos(α) < 0.999375.

Now, using the inverse cosine function:

α = cos⁻¹(0.999375).

Calculating this gives us α ≈ 8.1 degrees.

Final Thoughts

Thus, the maximum angle α that the sphere can be pushed aside without touching the floor is approximately 8.1 degrees. This analysis shows how the properties of materials, such as Young's modulus, play a crucial role in understanding the behavior of structures under load.