Question

A cowboy walks at a pace of about two steps per second, holding a glass of diameter \(10.0 \mathrm{~cm}\) that contains milk. The milk sloshes higher and higher in the glass until it eventually starts to spill over the top. Determine the maximum speed of the waves in the milk.

Short answer

Answer: The maximum speed of the waves in the milk when they start to spill over the top of the glass is 0.21 m/s.

Step-by-step solution

Identify the relevant properties of the glass and milk

Given that the diameter of the glass is \(10.0 \mathrm{~cm}\) and it contains milk, we can first find the radius of the glass by dividing the diameter by 2. This gives us: \(radius = \frac{diameter}{2} = \frac{10.0}{2} \mathrm{~cm} = 5.0 \mathrm{~cm}\) Next, let's understand the concept of wave speed in the milk. Wave speed depends on the properties of the liquid medium (in this case milk) and the radius of the circular container. The wave speed can be calculated using the following formula: \(wave\_speed = \sqrt{\frac{T}{\rho}} \cdot \frac{2 \pi}{k}\) where \(T\) is the surface tension of milk, \(\rho\) is the density of milk, and \(k\) is the wave number.

Calculate the wave speed

As the milk starts to spill over the top, the wave height in the glass would be equal to the radius of the glass (5.0 cm). Hence, we can use this value to find the wave number (k) using the following formula: \(k = \frac{2 \pi}{wavelength}\) The wavelength, in this case, can be considered twice the radius of the glass. So, we can write the formula for wave number as: \(k = \frac{2 \pi}{2 \cdot radius}\) Now, substituting the radius value of \(5.0 \mathrm{~cm}\), we get: \(k = \frac{2 \pi}{2 \cdot 5.0} = \frac{2 \pi}{10} \, \text{cm}^{-1}\) Next, we need to find the values of surface tension (\(T\)) and density (\(\rho\)) of milk. The average surface tension of milk is close to \(4.83 \times 10^{-2} \, \mathrm{N/m}\) and the density of milk is approximately \(1.03 \times 10^3 \, \mathrm{kg/m^{3}}\). Finally, we can calculate the wave speed using the formula mentioned in the previous step: \(wave\_speed = \sqrt{\frac{4.83 \times 10^{-2}}{1.03 \times 10^3}} \cdot \frac{2 \pi}{\frac{2 \pi}{10}}\) Simplifying and solving: \(wave\_speed = \sqrt{4.68 \times 10^{-5}} \cdot 10 = 0.21 \, \mathrm{m/s}\)

Find the maximum speed of the waves

The maximum speed of the waves in the milk when they start to spill over the top of the glass is: \(maximum\_ speed\_ of\_ waves = 0.21 \, \mathrm{m/s}\)