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Multiple-Concept Example 7 deals with the concepts that are important in this problem. A penny is placed at the outer edge of a disk (radius 0.150 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.80 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk.

Short Answer

Expert verified
The minimum coefficient of friction is approximately 0.186.

Step by step solution

01

Understand the Problem

We need to find the minimum coefficient of friction that allows the penny to stay on a rotating disk without slipping off. The disk has a radius of 0.150 m and a rotation period of 1.80 s.
02

Calculating the Angular Velocity

The angular velocity \( \omega \) can be calculated using the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period. Here, \( T = 1.80 \) s, so \[ \omega = \frac{2\pi}{1.80} \approx 3.49 \text{ rad/s}. \]
03

Determine the Centripetal Force Required

The centripetal force \( F_c \) required to keep the penny moving in a circle is given by \( F_c = m \omega^2 r \), where \( m \) is the mass of the penny, \( \omega \) is the angular velocity, and \( r \) is the radius of the disk. The radius is 0.150 m, and we've calculated \( \omega \) as 3.49 rad/s, so \[ F_c = m \times 3.49^2 \times 0.150. \]
04

Relate Centripetal Force to Frictional Force

For the penny not to slip, the frictional force \( F_f \) must be equal to or greater than the centripetal force. The frictional force is given by \( F_f = \mu mg \), where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
05

Solve for the Coefficient of Friction \( \mu \)

Set the frictional force equal to the centripetal force to solve for \( \mu \): \[ \mu mg = m \omega^2 r. \] Cancel \( m \) from both sides:\[ \mu = \frac{\omega^2 r}{g}. \]Substitute \( \omega = 3.49 \text{ rad/s} \), \( r = 0.150 \text{ m} \), and \( g = 9.81 \text{ m/s}^2 \):\[ \mu = \frac{(3.49)^2 \times 0.150}{9.81} \approx 0.186. \]

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Key Concepts

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

Angular Velocity
Angular velocity is a key concept in understanding rotational motion. Think of it as how quickly an object rotates around a central point or axis. Like speed describes how fast something moves in a straight line, angular velocity describes how fast something spins.
For a rotating object, angular velocity is calculated using the formula:
  • \( \omega = \frac{2\pi}{T} \)
where \( \omega \) is the angular velocity in radians per second, and \( T \) is the period of rotation in seconds.
In our problem, the disk completes one full rotation in 1.80 seconds. By plugging \( T = 1.80 \) into our formula, we find:
  • \( \omega = \frac{2\pi}{1.80} \)
  • \( \omega \approx 3.49 \text{ rad/s} \)
This means the disk spins at about 3.49 radians per second, which is a measure of how quickly all points on the disk, including where the penny is, are rotating.
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path. Imagine swinging a ball on a string; the tension in the string provides the centripetal force that keeps the ball moving in a circle rather than flying away.
For an object moving in a circle, the centripetal force is calculated using:
  • \( F_c = m \omega^2 r \)
where:
  • \( m \) is the mass of the object
  • \( \omega \) is the angular velocity
  • \( r \) is the radius of the circular path
In our problem, the penny must stay on the 0.150 m radius disk, which means it requires a centripetal force to maintain its circular motion. As the disk rotates, the penny experiences this force directed towards the center of the disk, ensuring it doesn’t slip off. The mass \( m \) of the penny isn’t needed for calculating the coefficient of friction, as it cancels out later in our problem.
Rotational Motion
Rotational motion refers to the motion of an object around a central point or axis. Unlike linear motion, where objects move from one point to another in a straight line, rotational motion involves turning around in a circular path.
In our exercise, the disk exhibits rotational motion by spinning around its center. As it rotates, each part of the disk, including where our penny is located, follows a circular path. This motion can be observed in everyday life, like the hands of a clock, wheels on a car, or Earth revolving around the Sun.
Understanding the forces involved in rotational motion is crucial because it allows us to predict how objects will behave when subjected to rotational forces. It’s important to know how fast the disk spins (angular velocity) and the force required to keep objects moving in a circle (centripetal force), to solve problems related to rotational motion effectively.
Disk Rotation
Disk rotation involves studying how an object, like a disk, rotates around its center. The rotating motion can be described by variables like the rotation period and angular velocity.
In our scenario, we are dealing with a disk that completes its rotation every 1.80 seconds. This rotation sets the stage for understanding how to keep the penny from slipping. As the disk rotates, the outer edge, where the penny is placed, is subject to forces that try to keep the penny on its circular path.
For the penny not to slip while the disk rotates, the friction between the penny and the disk must be enough to counteract the centripetal force pulling the penny outward. This is why we calculate the minimum coefficient of friction needed. The coefficient of friction represents how effectively the surfaces in contact resist slipping. Hence, understanding disk rotation helps in solving practical problems involving circular motion.

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