Problem 82 - 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 0.806.

Step 1: Understand the Problem

The cup starts to slip when the tray's amplitude of motion is 0.050 m and it moves with a frequency of 2.00 Hz. We need to find the coefficient of static friction between the tray and the cup.

Step 2: Find the Maximum Acceleration

The formula for acceleration in simple harmonic motion is given by \( a = -\omega^2 x \) where \( \omega \) is the angular frequency and \( x \) is the displacement. The maximum acceleration is when \( x \) is at the maximum amplitude, \( A \). Thus, \( a_{max} = \omega^2 A \).

Step 3: Calculate Angular Frequency

The angular frequency \( \omega \) is related to frequency \( f \) by \( \omega = 2\pi f \). Given \( f = 2.00 \) Hz, \( \omega = 2\pi \times 2.00 \) rad/s.

Step 4: Substitute for Maximum Acceleration

Using \( \omega = 4\pi \) rad/s, we substitute into the formula for maximum acceleration: \( a_{max} = (4\pi)^2 \times 0.050 \). Simplify to find \( a_{max} \).

Step 5: Relate Maximum Acceleration to Static Friction

The maximum acceleration is equal to the maximum static frictional force divided by the mass (\( a_{max} = \mu_s g \)). Equating the maximum acceleration to \( \mu_s g \), solve for \( \mu_s \).

Step 6: Solve for Static Friction Coefficient

We have \( a_{max} = \mu_s g \). Thus, \( \mu_s = \frac{a_{max}}{g} \) where \( g \approx 9.81 \ m/s^2 \). Substitute the value of \( a_{max} \) found.

Step 7: Calculate and Conclude

Substitute the values calculated to find \( \mu_s \). This will provide the coefficient of static friction between the tray and the cup.

Key Concepts

Simple Harmonic Motion

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement from the mean position.
It can be visualized as the motion of a pendulum or a mass attached to a spring. In the context of our problem, the tray moves back and forth in such motion. The motion can be described mathematically by sine or cosine functions. The sinusoidal nature of this motion involves characteristics such as amplitude, frequency, period, and phase.

• **Amplitude:** The maximum displacement from the equilibrium position, in this case, it's 0.050 m for the tray.
• **Frequency:** The number of oscillations per second. Given as 2.00 Hz, it tells us how fast the motion repeats itself.
• **Period:** The time for one complete cycle of motion, which can be found using the relationship with frequency, as period (\(T\)) is \(1/f\).
• **Phase:** Determines the position at a specific point in time. However, it doesn’t play a crucial role in this problem.

The oscillating tray affects the cup by applying variable forces due to its back-and-forth movement.

Angular Frequency

Angular frequency is a concept that describes how quickly an object moves along a circular path or in simple harmonic motion.
It is denoted as \(\omega\) and is calculated through the equation \(\omega = 2\pi f\), where \(f\) is the frequency of the motion.
In this problem, with a frequency of 2.00 Hz, the angular frequency is \(\omega = 2\pi \times 2.00 = 4\pi \) rad/s.
Angular frequency offers a convenient way to measure the rate of oscillation, particularly in radians per second, which offers a unified view across linear and rotational motions. It helps in analyzing the object's movement with respect to time, considering rotational aspects. Understanding angular frequency provides insight into how different elements such as speed and cycle time interact in oscillatory systems.

Maximum Acceleration

The maximum acceleration in simple harmonic motion occurs at the maximum displacement, or amplitude.
This is because the restoring force, hence acceleration, is greatest when the object is furthest from its balance point.
By using the formula for acceleration \(a = -\omega^2 x\), where \(x\) is the displacement, we find the maximum acceleration by substituting the amplitude. Given that \(x = A\), the formula becomes \(a_{max} = \omega^2 A\).
For our exercise, with \(\omega = 4\pi\) rad/s and \(A = 0.050\) m, the maximum acceleration is:\[a_{max} = (4\pi)^2 \times 0.050 \]
This calculation is critical as it relates to the maximum static friction that can be exerted before the cup begins to slip on the tray. Thus, it links the physical dynamics of the motion with the forces experienced by objects within that system.

Static Friction

Static friction is the force that keeps two objects in contact from sliding past each other when a force is applied. It acts to resist the start of motion.
The coefficient of static friction, \(\mu_s\), is a dimensionless value representing the ratio of the maximum static frictional force that can arise between the surfaces to the normal force.
In this case, when the cup starts slipping, the maximum acceleration from the tray equals the maximum static frictional force divided by the mass of the cup. This relationship is expressed as:\[a_{max} = \mu_s g \]
where \(a_{max}\) is the maximum acceleration calculated from the motion, and \(g\) is the acceleration due to gravity, approximately \(9.81 \ m/s^2\).
By rearranging this formula, we find:\[\mu_s = \frac{a_{max}}{g} \]The coefficient \(\mu_s\) is a key factor for determining whether an object will stay in place or start moving, important for calculations involving objects on moving surfaces.