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Chapter 5: Problem 20

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 \(=0.150 \mathrm{m}\) ) 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 0.187.

Step by step solution

01

Determine the velocity of the penny

To find the velocity of the penny, we first need to use the relation between the radius, the period of rotation, and the velocity. The formula for the speed of an object in circular motion is given by \( v = \frac{2\pi r}{T} \), where \( r = 0.150 \text{ m} \) is the radius and \( T = 1.80 \text{ s} \) is the period. Substituting these values, we have \( v = \frac{2\pi \times 0.150}{1.80} \approx 0.524 \text{ m/s} \).
02

Calculate the centripetal force required

The centripetal force required to keep the penny moving in a circle can be expressed as \( F_c = \frac{mv^2}{r} \). Since we don't have the mass of the penny, we'll denote it as \( m \). Thus, \( F_c = \frac{m \times (0.524)^2}{0.150} \approx 1.83m \text{ N} \).
03

Relate centripetal force to frictional force

The frictional force must provide the necessary centripetal force to keep the penny in circular motion. The frictional force \( F_f \) is given by \( F_f = \mu mg \), where \( \mu \) is the coefficient of friction and \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity. Equating the forces, we have \( \mu mg = 1.83m \).
04

Solve for the coefficient of friction

We can cancel the mass \( m \) from both sides of the equation because it appears on both sides. This gives us \( \mu g = 1.83 \). Solving for \( \mu \), we find \( \mu = \frac{1.83}{9.81} \approx 0.187 \).

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Key Concepts

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

Centripetal Force
When an object moves in a circle, it experiences a force known as centripetal force. This force acts towards the center of the circle, ensuring that the object continues its circular path. Without this force, the object would travel in a straight line due to inertia.
The magnitude of centripetal force can be calculated using the formula:
  • \( F_c = \frac{mv^2}{r} \)
Here:
  • \( F_c \) is the centripetal force,
  • \( m \) is the mass of the object,
  • \( v \) is the velocity of the object,
  • \( r \) is the radius of the circle.
This formula shows that centripetal force is directly related to the mass, velocity squared, and inversely to the radius. In our example, centripetal force keeps the penny on the rotating disk, pointing towards the center of the disk.
Coefficient of Friction
The coefficient of friction is a measure of how much frictional force exists between two surfaces. It is a dimensionless value that describes the ratio of the force of friction between two bodies to the force pressing them together.
There are two main types of friction:
  • Static Friction: Prevents movement between initially stationary surfaces.
  • Kinetic Friction: Occurs when surfaces slide against each other.
In this exercise, static friction is of interest because the penny does not slide off the rotating disk. The formula used to relate frictional force to the centrifugal force was \( F_f = \mu mg \), where:
  • \( F_f \) is frictional force,
  • \( \mu \) is the coefficient of friction,
  • \( m \) is the mass of the object (the penny),
  • \( g \) is the acceleration due to gravity.
By rearranging for \( \mu \), and knowing that frictional force must equal the centripetal force for the penny to stay on the disk, we find the minimum \( \mu \) needed.
Period of Rotation
The period of rotation is a fundamental concept that describes the time it takes for one complete cycle of motion. It is often denoted as \( T \) and is measured in seconds for rotational systems. In our example, the period is the time the disk takes to make one full rotation, which is 1.80 seconds.
Understanding the period helps in calculating the speed of the object in circular motion. The relationship between the period, radius, and velocity is given by the formula:
  • \( v = \frac{2\pi r}{T} \)
Where:
  • \( v \) is the velocity,
  • \( r \) is the radius of the circle,
  • \( T \) is the period of rotation.
This formula shows that the velocity and period are inversely related. A shorter period results in faster motion, as the object covers the same distance in less time. Calculating the velocity of the penny with this formula is a crucial step in finding the centripetal force and the necessary coefficient of friction.

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Most popular questions from this chapter

A satellite circles the earth in an orbit whose radius is twice the earth's radius. The earth's mass is \(5.98 \times 10^{24} \mathrm{kg}\), and its radius is \(6.38 \times 10^{6} \mathrm{m}\). What is the period of the satellite?

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be \(43.3 \mathrm{km} / \mathrm{s}\) and \(58.6 \mathrm{km} / \mathrm{s} .\) The slower planet's orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

The second hand and the minute hand on one type of clock are the same length. Find the ratio \(\left(a_{\text {c, second }} / a_{\text {c, minute }}\right)\) of the centripetal accelerations of the tips of the second hand and the minute hand.

A Dangerous Ride. You and your exploration team are stuck on a steep slope in the Andes Mountains in Argentina. A deadly winter storm is approaching and you must get down the mountain before the storm hits. Your path leads you around an extremely slippery, horizontal curve with a diameter of \(90 \mathrm{m}\) and banked at an angle of \(40.0^{\circ}\) relative to the horizontal. You get the idea to unpack the toboggan that you have been using to haul supplies, load your team upon it, and ride it down the mountain to get enough speed to get around the banked curve. You must be extremely careful, however, not to slide down the bank: At the bottom of the curve is a steep cliff. (a) Neglecting friction and air resistance, what must be the speed of your toboggan in order to get around the curve without sliding up or down its bank? Express your answer in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{m} .\) p.h. (b) You will need to climb up the mountain and ride the toboggan down in order to attain the speed you need to safely navigate the curve (from part (a)), The mountain slope leading into the curve is at an angle of \(30.0^{\circ}\) relative to the horizontal, and the coefficient of kinetic friction between the toboggan and the surface of the slope is \(\left(\mu_{\text {mrantain }}=0.15\right) .\) How far up the mountain (distance along the slope, not elevation) from the curve should you start your ride? Note: the path down the mountain levels off at the bottom so that the toboggan enters the curve moving in the horizontal plane (i.e., in the same plane as the curve).

A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of \(28 \mathrm{m} / \mathrm{s},\) the block hangs vertically down. But when the van maintains this same speed around an unbanked curve (radius \(=150 \mathrm{m}),\) the block swings toward the outside of the curve. Then the string makes an angle \(\theta\) with the vertical. Find \(\theta\).
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