/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 A small button placed on a horiz... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 48

A small button placed on a horizontal rotating platform with diameter 0.320 \(\mathrm{m}\) will revolve with the platform when it is brought up to a speed of 40.0 rev/min, provided the button is no more than 0.150 \(\mathrm{m}\) from the axis. (a) What is the coefficient of static friction between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 \(\mathrm{rev} / \mathrm{min}\) ?

Short Answer

Expert verified
(a) 0.267, (b) 0.066 m

Step by step solution

01

Convert Rotational Speed to Radians per Second

First, convert the rotational speed from revolutions per minute to radians per second. Remember that one revolution is equal to \(2\pi\) radians. For 40.0 revolutions per minute, we calculate: \[ \text{Angular velocity} \ (\omega) = \frac{40.0 \times 2\pi \ \text{radians}}{60 \ \text{seconds}} \approx 4.19 \ \text{rad/s} \]
02

Calculate the Maximum Static Frictional Force

Using the formula for centripetal force \(F_c = m\omega^2 r\), where \(m\) is mass, \(\omega\) is angular velocity, and \(r\) is the radius from the axis, we determine the maximum static frictional force: The static frictional force that prevents slipping is \(f_s = \mu_s m g\). Equating the centripetal force and the frictional force gives \(m\omega^2 r = \mu_s mg\) and thus:\[ \mu_s = \frac{\omega^2 r}{g} \]
03

Apply Values to Find the Coefficient of Static Friction (Part a)

Plug the values of \(\omega\), \(r = 0.150 \ m\), and \(g = 9.81 \ m/s^2\) into the equation:\[ \mu_s = \frac{(4.19 \ \text{rad/s})^2 \times 0.150 \ m}{9.81 \ m/s^2} \approx 0.267 \]
04

Determine New Angular Velocity for 60 rev/min

Convert 60.0 revolutions per minute to radians per second: \[ \omega_{new} = \frac{60.0 \times 2\pi \ \text{radians}}{60 \ \text{seconds}} = 6.28 \ \text{rad/s} \]
05

Calculate Maximum Radius for New Speed (Part b)

Using the calculated coefficient of static friction \(\mu_s = 0.267\), find the new potential radius where friction still balances the centripetal force: \[ \mu_s mg = m\omega_{new}^2 r_{new} \]Solving for \(r_{new}\), plug in the values:\[ r_{new} = \frac{\mu_s g}{\omega_{new}^2} = \frac{0.267 \times 9.81}{(6.28)^2} \approx 0.066 \ m \]

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Key Concepts

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

Centripetal Force
Centripetal force is a crucial concept when dealing with objects in circular motion. It's the force that keeps an object moving in a circle and prevents it from flying outwards. This force acts towards the center of the circle, hence its name: 'centripetal,' which means 'center-seeking'.

Whenever an object is in circular motion, like the button on the rotating platform, it needs a force to keep it turning. This force can come from various sources, such as tension, gravity, or, as in this case, friction. The equation for centripetal force is given by:
  • \( F_c = m\omega^2r \)
where:
  • \( F_c \) is the centripetal force,
  • \( m \) is the mass of the object,
  • \( \omega \) is the angular velocity, and
  • \( r \) is the radius from the center of rotation.
Understanding centripetal force is essential since it's the force that the static friction needs to overcome to prevent the button from sliding off the platform.
Angular Velocity
Angular velocity is another key aspect of circular motion. It's a measure of how fast an object rotates or revolves around a central point. Angular velocity is represented as \( \omega \) and is typically measured in radians per second.

To calculate angular velocity, you convert the rotational speed from revolutions per minute (rev/min) to radians per second (rad/s) using the formula:
  • \( \omega = \frac{N \times 2\pi}{60} \)
where \( N \) is the number of revolutions per minute.

In the exercise, converting 40 rev/min resulted in an angular velocity of approximately 4.19 rad/s, which was necessary for determining how much static friction was needed to keep the button from slipping. Similarly, the exercise required a new calculation with 60 rev/min, resulting in 6.28 rad/s.
Coefficient of Static Friction
Static friction is the force that prevents two surfaces from sliding past one another. In this exercise, the static friction between the button and the rotating platform is what prevents the button from sliding off due to rotating motion. The 'coefficient of static friction,' denoted as \( \mu_s \), is a dimensionless value that represents how much frictional force exists between the two surfaces.

The coefficient is essential in determining how much centripetal force can be provided by friction. You find it using the relationship:
  • \( \mu_s = \frac{\omega^2 r}{g} \)
where \( g \) is the acceleration due to gravity, approximately 9.81 m/s虏.

In this case, solving for \( \mu_s \) using the values \( \omega = 4.19 \text{ rad/s} \) and \( r = 0.150 \text{ m} \) gives a coefficient close to 0.267. This number is crucial for determining whether the button remains stationary on the platform during rotation. With a change in the rotational speed, the necessary coefficient to prevent slipping is recalculated to find the furthest distance the button can be placed from the center.

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Most popular questions from this chapter

You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

A steel washer is suspended inside an empty shipping crate from a light string attached to the top of the crate. The crate slides down a long ramp that is inclined at an angle of \(37^{\circ}\) above the horizontal. The crate has mass 180 \(\mathrm{kg}\) . You are sitting inside the crate (with a flashlight); your mass is 55 \(\mathrm{kg}\) . As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate when the string makes an angle of \(68^{\circ}\) with the top of the crate. What is the coefficient of kinetic friction between the ramp and the crate?

If the coefficient of static friction between a table and a uniform massive rope is \(\mu_{s},\) what fraction of the rope can hang over the edge of the table without the rope sliding?

An airplane flies in a loop (a circular path in a vertical plane) of radius 150 \(\mathrm{m} .\) The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom. (a) At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point? (b) At the bottom of the loop, the speed of the airplane is 280 \(\mathrm{km} / \mathrm{h}\) . What is the apparent weight of the pilot at this point? His true weight is 700 \(\mathrm{N}\) .

A 40.0 -kg packing case is initially at rest on the floor of a 1500 -kg pickup truck. The coefficient of static friction between the case and the truck floor is \(0.30,\) and the coefficient of kinetic friction is \(0.20 .\) Before each acceleration given below, the truck is traveling due north at constant speed. Find the magnitude and direction of the friction force acting on the case (a) when the truck accelerates at 2.20 \(\mathrm{m} / \mathrm{s}^{2}\) northward and (b) when it accelerates at 3.40 \(\mathrm{m} / \mathrm{s}^{2}\) southward.
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