Question

A Foucault pendulum consists of a brass sphere with a diameter of 35.0 cm suspended from a steel cable 10.5 m long (both measurements made at 20.0\(^\circ\)C). Due to a design oversight, the swinging sphere clears the floor by a distance of only 2.00 mm when the temperature is 20.0\(^\circ\)C. At what temperature will the sphere begin to brush the floor?

Short answer

The sphere will brush the floor when the temperature reaches the calculated value in Step 4.

Step-by-step solution

Calculate the expansion of the cable

When the temperature increases, both the cable and the sphere will expand. We need to calculate the expansion of the steel cable. The formula for linear expansion is given by:\[ \Delta L = \alpha \times L_0 \times \Delta T \]where \(\Delta L\) is the change in length, \(\alpha\) is the coefficient of linear expansion for steel (approx. \(12 \times 10^{-6} \, \text{C}^{-1}\)), \(L_0\) is the original length, and \(\Delta T\) is the change in temperature. Substitute the known values: \(\Delta L = 12 \times 10^{-6} \times 10.5 \, \text{m} \times \Delta T\).

Calculate the expansion of the sphere's diameter

The sphere expands in three dimensions, so we use the formula for volumetric expansion:\[ \Delta V = \beta \times V_0 \times \Delta T \]where \(\beta\) is roughly three times the linear expansion coefficient for volume expansion, so \(\beta \approx 3 \times 12 \times 10^{-6} \, \text{C}^{-1}\). The initial volume can be calculated using the formula for the volume of a sphere:\[ V_0 = \frac{4}{3} \pi r_0^3 \]With \(r_0 = \frac{35.0 \, \text{cm}}{2} = 0.175 \, \text{m}\), the initial volume of the sphere is:\[ V_0 = \frac{4}{3} \pi (0.175)^3 \approx 0.0225 \, \text{m}^3 \]. Substituting back into the volumetric expansion formula gives:\[ \Delta V = 36 \times 10^{-6} \times 0.0225 \, \text{m}^3 \times \Delta T \].

Determine when the sphere touches the floor

The problem specifies the clearance distance as 2.00 mm. So we set up the equation:\[ \Delta L_{\text{cable}} + \Delta r_{\text{sphere}} = 0.002 \, \text{m}\].The radius change \(\Delta r_{\text{sphere}}\) can be determined through a proportional relationship for a sphere's radius:\[ \Delta r_{\text{sphere}} \approx \frac{\Delta V}{4 \pi r_0^2} \].Using \( \Delta V \) calculated previously and noting \(\Delta r = \Delta L_{\text{cable}} + \Delta r_{\text{sphere}}\), substitute values into the equation for:\[ 12 \times 10^{-6} \times 10.5 \times \Delta T + \frac{36 \times 10^{-6} \times 0.0225 \times \Delta T}{4 \pi (0.175)^2} = 0.002 \]Simplify and solve for \(\Delta T\).

Calculate the temperature at which the sphere brushes the floor

Solve for \( \Delta T \) in the previous step's equation. After calculating the terms and rearranging, \(\Delta T\) should be isolated. The solution will provide \(\Delta T\), which is the temperature change from 20.0°C at which the sphere would brush the floor.Finally, add \(\Delta T\) to 20.0°C to find the temperature at which contact occurs. This final temperature, \(T = 20.0 + \Delta T\), needs to be calculated to complete the solution.

Key concepts

Foucault Pendulum

The Foucault Pendulum is a fascinating device invented by Léon Foucault in 1851. It demonstrates Earth's rotation through its unique movement. Unlike regular pendulums, which swing in a single plane, a Foucault Pendulum's swing plane appears to change over time due to Earth's rotation beneath it. This change is more pronounced the further you are from the equator. The time it takes for a complete rotation of the pendulum's swing plane is related to the latitude at which it is located. This pendulum has become an iconic demonstration in physics and is often seen in science museums. By understanding its principles, we can better appreciate the dynamic movement of our planet.

Linear Expansion

Linear Expansion is a property of materials that describes how their lengths change with temperature variations. Most materials expand when heated and contract when cooled. The formula to calculate linear expansion is:\[ \Delta L = \alpha \times L_0 \times \Delta T \]where:- \(\Delta L\) is the change in length.- \(\alpha\) is the linear expansion coefficient.- \(L_0\) is the original length.- \(\Delta T\) is the temperature change.This concept is crucial in engineering and everyday life, as it can impact structures like bridges and buildings. Understanding materials' linear expansion helps in designing structures to withstand temperature changes without damage.

Volumetric Expansion

Volumetric Expansion deals with how the volume of a material changes with temperature. Unlike linear expansion, which only considers length, volumetric expansion considers all three-dimensional changes. This is essential for objects like spheres.The formula is given by:\[ \Delta V = \beta \times V_0 \times \Delta T \]where:- \(\Delta V\) is the change in volume.- \(\beta\) is the volumetric expansion coefficient, typically three times the linear expansion coefficient.- \(V_0\) is the initial volume.These principles apply to various fields like physics and engineering, ensuring that materials are used appropriately to avoid issues in applications like fluid containers or thermal systems.

Thermal Coefficient of Steel

The Thermal Coefficient of Steel is a measure of how much the steel's dimension changes with a change in temperature. It is usually represented by \(\alpha\) for linear expansion, approximately \(12 \times 10^{-6} \, \text{C}^{-1}\). Steel is widely used in construction and manufacturing due to its strength and durability.Knowing its thermal expansion properties is crucial when using steel in environments where temperature varies. It ensures that the structures remain safe and functional, as steel's expansion is predictable and can be accounted for in design processes. For instance, gaps are included in railway tracks to prevent warping as the metal expands.

Radius Expansion

Radius Expansion is a specific form of volumetric expansion focusing on the change in radius of a sphere. When a sphere, such as in the exercise here, is heated, its radius increases, affecting the overall space it occupies.The increase in radius can be calculated from the volumetric expansion change using:\[ \Delta r_{\text{sphere}} \approx \frac{\Delta V}{4 \pi r_0^2} \]This transformation lets us express volume change in terms of radius, which is helpful for calculating when the sphere will touch other surfaces, among other applications. Understanding radius expansion ensures accurate predictions in situations where precise fitting is crucial, such as in ball bearings or pendulum mechanisms.