A penny is sitting on the edge of an old phonograph disk that is spinning at 33 rpm and has a diameter of 12 inches. What is the minimum coefficient of static friction between the penny and the surface of the disk to ensure that the penny doesn't fly off?

Short Answer

Expert verified
Answer: The minimum coefficient of static friction is approximately 0.155.

Step by step solution

01

Calculate the angular speed of the disk

First, we need to find the angular speed of the disk in radians per second. Given the spinning rate of 33 rpm (rotations per minute), we can convert it to radians per second using the conversion factor (2 * pi radians per rotation) and the fact that there are 60 seconds per minute: Angular speed (ω) = (33 rpm * 2π rad/rotation) / 60 sec/min
02

Calculate the radius of the disk

We are given the diameter of the disk, which is 12 inches. To find the radius, we can simply divide the diameter by 2: Radius (r) = Diameter / 2 Radius (r) = 12 inches / 2 Radius (r) = 6 inches
03

Calculate the centripetal force acting on the penny

To calculate the centripetal force (Fc) acting on the penny, we can use the formula for centripetal force: Fc = m * (ω^2) * r where m is the mass of the penny. However, we do not need the exact value of Fc, because we are looking for the ratio between the forces (coefficient of static friction), so we leave the centripetal force as a function of mass: Fc = m * (ω^2) * r
04

Calculate the force of static friction

The force of static friction (Fs) that keeps the penny from flying off the phonograph disk can be calculated using the formula: Fs = μ * Fn where μ is the coefficient of static friction and Fn is the normal force. Since the penny is on a horizontal surface, the normal force will be equal to the gravitational force acting on the penny (Fn = m * g), where g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, we have: Fs = μ * m * g
05

Find the minimum coefficient of static friction

To ensure that the penny doesn't fly off, the force of static friction (Fs) must be greater than or equal to the centripetal force (Fc): Fs >= Fc μ * m * g >= m * (ω^2) * r Since the mass of the penny (m) appears on both sides of the equation, we can divide both sides by m: μ * g >= (ω^2) * r Now, we can solve for the minimum coefficient of static friction (μ) by dividing both sides by g: μ >= ((ω^2) * r) / g Plugging in the values we calculated earlier for ω, r, and g (Note: Make sure to convert inches to meters for consistency in units): ω = (33 * 2π) / 60 = π rad/s r = 6 inches * 0.0254 m/inch ≈ 0.1524 m g = 9.8 m/s^2 μ >= ((π^2) * 0.1524) / 9.8 μ >= 0.155 Thus, the minimum coefficient of static friction between the penny and the surface of the disk to ensure that the penny doesn't fly off is approximately 0.155.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
Understanding centripetal force is crucial when analyzing the motion of objects moving in a circle. In our example of a penny on a spinning phonograph disk, centripetal force is the invisible 'hand' that keeps the penny from flying off. It's directed towards the center of the circular path and is calculated with the formula:
\( F_c = m \times \omega^2 \times r \), where \( F_c \) represents the centripetal force, \( m \) is the mass of the object (in this case, the penny), \( \omega \) is the angular speed, and \( r \) is the radius. For the penny to stay put, the static friction force must be equal to or greater than the centripetal force needed to maintain its circular motion.

This concept is crucial for things like designing roads with curves, amusement park rides, and in this particular case, understanding why the penny stays on the spinning record. It helps us grasp the relationship between the rotational velocity of an object and the force required to keep it moving in a loop without losing grip.
Angular Speed
The angular speed of an object, denoted by the symbol \( \omega \), is a measure of how fast it rotates or revolves around a central point. In the given exercise, we calculated the angular speed of the phonograph disk by converting rotations per minute (RPM) to radians per second. This is done with the conversion factor \( 2\pi \) radians for every rotation and factoring in the 60 seconds per minute.
\[ \omega = \left( \frac{33 \times 2\pi}{60} \right) \text{ rad/s} \].

Angular speed helps us understand how different parts of an object move at different speeds. For example, the edge of a spinning record moves faster than a point closer to the center. By accounting for angular speed in physics problems, we can better predict the rotational behavior of objects and ensure they operate safely within the limits of the forces acting upon them.
Physics Problem Solving
Problem solving in physics often involves applying conceptual understanding to concrete situations, as exemplified by the penny and phonograph disk problem. Successful solving begins with identifying the correct principles, like Newton's laws, friction, and circular motion, and then using appropriate formulas to find the solution. The process we used:
  • Identify the problem: preventing the penny from flying off the spinning disk.
  • Understanding relevant concepts: angular speed, static friction, centripetal force.
  • Using formulas to express the physical quantities: \( F_c = m \times \omega^2 \times r \) and \( F_s = \mu \times m \times g \).
  • Combining and manipulating these formulas to isolate the variable of interest: \( \mu \).
  • Substituting the numbers and solving.
Through a structured approach to the problem, the elegant relationship between the forces at play was revealed, leading to the safe spinning of the penny which illustrates the beauty and efficacy of physics problem solving.

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Most popular questions from this chapter

A typical Major League fastball is thrown at approximately \(88 \mathrm{mph}\) and with a spin rate of \(110 \mathrm{rpm} .\) If the distance between the pitcher's point of release and the catcher's glove is exactly \(60.5 \mathrm{ft},\) how many full turns does the ball make between release and catch? Neglect any effect of gravity or air resistance on the ball's flight.

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