Question

A Foucault pendulum consists of a brass sphere with a diameter of \({\bf{35}}.{\bf{0}}{\rm{ }}{\bf{cm}}\) suspended from a steel cable \({\bf{10}}.{\bf{5}}{\rm{ }}{\bf{m}}\)long (both measurements made at \({\bf{20}}{\bf{.0}}\;{\bf{^\circ C}}\)). Due to a design oversight, the swinging sphere clears the floor by a distance of only \({\bf{2}}{\bf{.00}}\;{\bf{mm}}\) when the temperature is \({\bf{20}}{\bf{.0}}\;{\bf{^\circ C}}\). At what temperature will the sphere begin to brush the floor?

Short answer

The steel sphere will brush the floor at \(35.9\;^\circ {\rm{C}}\).

Step-by-step solution

Given Data

Diameter of the sphere:\(d = 35\;{\rm{cm}}\)

Length of steel cable:\(L = 10.5\;{\rm{m}}\)

The temperature at which measurements are made:\(T = 20.0\;^\circ {\rm{C}} = 293\;{\rm{K}}\)

The expansion in length:\(\Delta L = 2.0\;{\rm{mm}}\)

The coefficient of expansion of steel is: \(1.2 \times {10^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}\)

Concept

It is a common observation that matter changes its shape, size, volume, or density in response to variation in temperature. This phenomenon is known as thermal expansion. If the length of any object changes in response to the variation in temperature, it is called Linear Expansion. The increase in length \(\left( {{\bf{\Delta L}}} \right)\) can be calculated as-

\({\bf{\Delta L = \alpha L\Delta T}}\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Here, \({\bf{\alpha }}\) is the coefficient of expansion and \({\bf{\Delta T}}\) is the variation in temperature.

Calculation

Using equation (1), the increase in length is given as-

\(\begin{array}{c}\Delta L = \alpha L\Delta T\\ = \alpha L\left( {{T_o} - T} \right)\end{array}\)

Here\({T_o}\)is the final temperature at which sphere will brush the floor.

For the given values the equation becomes-

\(\begin{array}{l}2 \times {10^{ - 3}}\;{\rm{m}}\; = \left( {1.2 \times {{10}^{ - 5}}\;{{\rm{K}}^{{\rm{ - 1}}}}} \right)\left( {10.5\;{\rm{m}}} \right)\left( {{T_o} - 293\;{\rm{K}}} \right)\\\left( {{T_o} - 293\;{\rm{K}}} \right) = 15.9\;{\rm{K}}\end{array}\)

Solving further,

\(\begin{array}{c}{T_o} = \left( {293 + 15.9} \right)\;{\rm{K}}\\ = 308.9\;{\rm{K}}\\ = 35.9\;^\circ {\rm{C}}\end{array}\)

The temperature at which the sphere will brush the surface is \(35.9\;^\circ {\rm{C}}\).