Problem 42 - University Physics with Modern Physics

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.

Step-by-Step Solution

Answer

Answer: The maximum speed of the waves in the milk is approximately \(1.2\pi\,\text{m/s}\).

Step 1: Calculate the maximum acceleration

As the cowboy takes two steps per second, we can determine the frequency of the steps. \(frequency = 2 \,\text{steps/second}\) And the time period of each step can be calculated as the inverse of the frequency. \(T = 1 \div 2 = 0.5 \,\text{seconds/step}\) Since the cowboy walks at a constant pace, we can assume that the movement resembles simple harmonic motion (SHM). Therefore, the acceleration (horizontal) for any simple harmonic motion is given by: \(a = A\omega^{2}\) Where A is the amplitude, and ω is the angular frequency. As the amplitude is equal to the diameter of the glass \(A = 10.0\,\text{cm} = 0.1\,\text{m}\) Now, we need to find the angular frequency ω using the time period. \(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.5} = 4\pi\,\text{rad/s}\) Now, we can calculate the acceleration of the milk. \(a = A\omega^{2} = 0.1 \cdot (4\pi)^{2} = 16 \pi^{2}\,\text{m/s}^2\) Step 2: Calculate the maximum speed of the waves

Step 2: Use the conservation of mechanical energy

Since the milk waves can be assumed to undergo SHM, the conservation of mechanical energy can be used to find the maximum speed of the waves. Initially, the complete energy is potential, and as the wave moves down, the potential energy transforms into kinetic energy. The potential energy (EP) can be calculated as: \(EP = \frac{1}{2}m \cdot a \cdot A = \frac{1}{2}mA^{2}\omega^{2}\) At the maximum speed, the kinetic energy (EK) is equal to the potential energy. Therefore, we can write the equation as: \(EK = \frac{1}{2}mv^{2} = \frac{1}{2}mA^{2}\omega^{2}\) where m is the mass of milk involved, v is the maximum speed, A is the amplitude, and ω is the angular frequency. The mass m will cancel out from both sides of the equation. \(v^{2} = A^{2}\omega^{2}\) \(v = \sqrt{A^{2}\omega^{2}}= A\omega = 0.1\,\text{m} \cdot 4\pi\,\text{rad/s} = 1.2\pi\,\text{m/s}\) Thus the maximum speed of the waves in the milk is approximately \(1.2\pi\,\text{m/s}\).