Problem 83 - Physics

Question

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{~Hz}\). On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{~m}\).

Step-by-Step Solution

Answer

The coefficient of static friction is approximately 0.805.

Step 1: Understanding Simple Harmonic Motion

The motion of the tray can be described using simple harmonic motion (SHM), where the displacement is given by \( x(t) = A \cos(\omega t) \). The angular frequency \( \omega \) is related to the frequency by \( \omega = 2 \pi f \). Here, \( f = 2.00 \mathrm{~Hz} \).

Step 2: Calculating Angular Frequency

Calculate the angular frequency \( \omega \) using the formula: \( \omega = 2 \pi f \). Substituting \( f = 2.00 \mathrm{~Hz} \), we get \( \omega = 2 \pi \times 2.00 = 4\pi \mathrm{~rad/s} \).

Step 3: Finding Maximum Acceleration

In SHM, the maximum acceleration \( a_{\text{max}} \) is given by \( a_{\text{max}} = A \omega^2 \). Substitute the values \( A = 5.00 \times 10^{-2} \mathrm{~m} \) and \( \omega = 4\pi \mathrm{~rad/s} \): \( a_{\text{max}} = (5.00 \times 10^{-2})(4\pi)^2 \).

Step 4: Solving for Maximum Acceleration

Calculate \( a_{\text{max}} = 5.00 \times 10^{-2} \times 16\pi^2 = 5.00 \times 10^{-2} \times 157.92 \approx 7.90 \mathrm{~m/s^2} \).

Step 5: Understanding Static Friction Force

The maximum static friction force \( f_s \) is given by \( f_s = \mu_s mg = ma_{\text{max}} \) where \( \mu_s \) is the static friction coefficient and \( m \) is the mass of the cup. The cup begins to slip when \( a_{\text{max}} \) exceeds \( g \mu_s \).

Step 6: Calculating Static Friction Coefficient

Use the relation \( a_{\text{max}} = g \mu_s \), where \( g = 9.81 \mathrm{~m/s^2} \). Rearrange to find \( \mu_s \): \[ \mu_s = \frac{a_{\text{max}}}{g} \]. Substitute \( a_{\text{max}} = 7.90 \mathrm{~m/s^2} \) and \( g = 9.81 \mathrm{~m/s^2} \): \[ \mu_s = \frac{7.90}{9.81} \approx 0.805 \].

Key Concepts

Angular Frequency

The concept of angular frequency is pivotal in understanding simple harmonic motion (SHM). In SHM, objects oscillate back and forth in a regular pattern, such as the tray moving horizontally in the exercise. Angular frequency, often denoted as \( \omega \), describes how rapidly an object completes one cycle of its motion. It is given by the formula \( \omega = 2 \pi f \), where \( f \) is the frequency of the motion.- For example, if an object oscillates with a frequency of \( 2.00 \mathrm{~Hz} \), meaning it completes 2 cycles per second, the angular frequency becomes \( 4\pi \mathrm{~rad/s} \).Angular frequency is expressed in radians per second (rad/s), as it relates to the angle covered by the oscillating object in its cycle. Understanding angular frequency helps in determining other critical aspects of the motion, like its speed and acceleration.

Amplitude

Amplitude is one of the key characteristics of simple harmonic motion, representing the maximum extent of the oscillation. It measures how far the system moves from its central or equilibrium position in either direction.- In our exercise, the amplitude is given as \( 5.00 \times 10^{-2} \mathrm{~m} \), or 0.05 meters.- The amplitude is a measure of energy in the system; a larger amplitude means more energy.In simple harmonic motion formulas, the amplitude, denoted \( A \), is used to calculate maximum displacement, speed, and in particular, the maximum acceleration. It is vital in describing the scale and energy of the motion involved.

Static Friction

Static friction is the force that prevents two surfaces from slipping past one another when in contact. It plays a crucial role when an object is at rest relative to another surface and begins to move.- The static friction force can be calculated using \( f_s = \mu_s mg \), where \( \mu_s \) is the static friction coefficient, \( m \) is the mass, and \( g \) is gravitational acceleration.In the context of this exercise, understanding static friction is important as it determines the point at which the cup starts sliding off the tray. This is when the tray's acceleration exceeds the static friction's capability to keep the cup in position without slipping.- The coefficient of static friction is found by equating the maximum acceleration with gravitational force multiplied by the coefficient of static friction, leading to \( \mu_s = \frac{a_{\text{max}}}{g} \).

Maximum Acceleration

The concept of maximum acceleration is critical in understanding how fast an object changes its velocity at its peak motion in simple harmonic motion. In the exercise, maximum acceleration is determined by both the amplitude of the motion and its angular frequency.- The maximum acceleration \( a_{\text{max}} \) in SHM is given by the formula \( a_{\text{max}} = A \omega^2 \).Using the amplitude \( A = 5.00 \times 10^{-2} \mathrm{~m} \) and the angular frequency \( \omega = 4\pi \mathrm{~rad/s} \) calculated earlier, \( a_{\text{max}} \) can be determined as approximately \( 7.90 \mathrm{~m/s^2} \).This acceleration is significant, as it needs to be lower than the static frictional force in order to keep the cup from sliding. If the tray accelerates beyond this limit, it causes the cup to slip, as indicated by static friction value calculations.