/*! 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 20 A coin rests \(15.0 \mathrm{~cm}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 20

A coin rests \(15.0 \mathrm{~cm}\) from the center of a turntable. The coefficient of static friction between the coin and turntable surface is \(0.350\). The turntable starts from rest at \(t=0\) and rotates with a constant angular acceleration of \(0.730 \mathrm{rad} / \mathrm{s}^{2}\). (a) Once the turntable starts to rotate, what force causes the centripetal acceleration when the coin is stationary relative to the turntable? Under what condition does the coin begin to move relative to the turntable? (b) After what period of time will the coin start to slip on the turntable?

Short Answer

Expert verified
a) Once the turntable starts to rotate, the centripetal force, which is caused by static friction, causes the coin to accelerate. The coin begins to move when the centripetal force exceeds the force of static friction. b) The coin starts to slip after a time period \( t = \sqrt{\mu_s g / r} / \alpha \)

Step by step solution

01

Determine the static friction force

The static friction force can be calculated by the equation \( F_{s} = \mu_s F_{n} \), where \( \mu_s \) is the coefficient of static friction and \( F_{n} \) is the normal force. In this case, \( F_{n} \) is the gravity acting on the coin, \( F_{n} = m g \), where \( m \) is the mass of the coin and \( g \) is the acceleration due to gravity. However, the mass of the coin is not given in the problem. We can say the static friction force \( F_{s} = \mu_s m g \).
02

Find the centripetal force

The centripetal force \( F_{c} \) is needed to keep the coin moving in a circle. It is given by \( F_{c} = m a_{c} \), where \( m \) is the mass of the coin and \( a_{c} \) is the centripetal acceleration. The centripetal acceleration \( a_{c} \) can be calculated using the formula \( a_{c} = r \omega^{2} \), where \( r \) is the radial distance from the center and \( \omega \) is the angular velocity. At the point of starting to slip, the angular velocity \( \omega \) reaches a maximum value, making \( F_{c} = \mu_s m g \). Therefore, \( r \omega^{2} = \mu_s g \)
03

Determine the angular velocity at which the coin starts to slip

The equation from Step 2 can be rearranged to solve for \( \omega \), the angular velocity at which the coin starts to slip: \( \omega = \sqrt{\mu_s g / r} \)
04

Find the time it takes for the coin to start to slip

The turntable has a constant angular acceleration \( \alpha \), so its angular velocity at any time can be calculated by \( \omega = \alpha t \). We know the angular velocity at which the coin starts to slip \( \omega_{s} \), so the time it takes for the coin to start to slip is \( t = \omega_{s} / \alpha \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Static Friction
Static friction plays a crucial role in keeping objects stationary relative to one another. In our case, it helps to keep the coin in place on the rotating turntable. Static friction is the force that must be overcome to start moving an object that is at rest. It's calculable using the formula:
  • \( F_s = \mu_s F_n \)
Here, \( F_s \) is the static friction force, \( \mu_s \) is the coefficient of static friction, and \( F_n \) is the normal force. For a horizontal surface, the normal force is typically the gravitational pull on the object, \( F_n = mg \), where \( m \) is the mass and \( g \) is the gravitational acceleration.
When the turntable starts spinning, the static friction must provide enough force to keep the coin stationary relative to the turntable. If the force required to maintain this state exceeds the static friction, the coin will begin to slide. The condition for the coin to start moving is when the centripetal force needed equals the maximum static friction available.
Angular Acceleration
Angular acceleration is a concept that describes how quickly an object's rotational speed changes over time. For the turntable, it refers to how fast it moves from being at rest to a certain rotational speed. It is denoted by \( \alpha \) and measured in radians per second squared.When we have a turntable that starts from a standstill, the constant angular acceleration formula tells us:
  • \( \omega = \alpha t \)
This equation shows that the angular velocity \( \omega \) at any given time \( t \) is directly proportional to the angular acceleration \( \alpha \).
In our problem, this formula helps us determine when the coin will start slipping. Once the angular velocity \( \omega \) reaches a certain point where static friction can no longer provide the necessary force for centripetal acceleration, slipping occurs. Thus, angular acceleration is essential in calculating how long it takes for this critical point to be reached.
Centripetal Acceleration
Centripetal acceleration is the acceleration required to make an object follow a curved path. In our turntable scenario, it's the acceleration directing the coin towards the center, ensuring it stays in circular motion.The equation for centripetal acceleration is:
  • \( a_c = r \omega^2 \)
where \( r \) is the radius or distance to the center, and \( \omega \) is the angular velocity. The turntable imposes this centripetal acceleration on the coin as it spins.
For the coin to remain stationary on the platform and rotate with the turntable, static friction must equal the centripetal force required. When the centripetal force \( F_c = m a_c \) surpasses the static friction capacity, the coin will start moving outward, slipping off the turntable. Calculating the critical value of \( \omega \) when this happens allows us to find the time when the coin will slip, using the relation between angular velocity and time from angular acceleration.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(0.400-\mathrm{kg}\) pendulum bob passes through the lowest part of its path at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). (a) What is the tension in the pendulum cable at this point if the pendulum is \(80.0 \mathrm{~cm}\) long? (b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (c) What is the tension in the pendulum cable when the pendulum reaches its highest point?

A stuntman whose mass is \(70 \mathrm{~kg}\) swings from the end of a \(4.0-\mathrm{m}-\mathrm{long}\) rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions in the rope that are required to make him follow his circular path (a) at the beginning of his motion, (b) at a height of \(1.5 \mathrm{~m}\) above the bottom of the circular arc, and (c) at the bottom of the arc.

A satellite of Mars, called Phoebus, has an orbital radius of \(9.4 \times 10^{6} \mathrm{~m}\) and a period of \(2.8 \times 10^{4} \mathrm{~s}\). Assuming the orbit is circular, determine the mass of Mars.

An air puck of mass \(m_{1}=0.25 \mathrm{~kg}\) is tied to a string and allowed to revolve in a circle of radius \(R=1.0 \mathrm{~m}\) on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of \(m_{2}=1.0 \mathrm{~kg}\) is tied to it (Fig. P7.27). The suspended mass remains in equilibrium while the puck on the tabletop revolves. (a) What is the tension in the string? (b) What is the horizontal force acting on the puck? (c) What is the speed of the puck? the horizontal force acting on the puck? (c) What is the the horizontal force acting on the puck? (c) What is the speed of the puck?

Because of Earth's rotation about its axis, a point on the equator has a centripetal acceleration of \(0.0340 \mathrm{~m} / \mathrm{s}^{2}\), whereas a point at the poles has no centripetal acceleration. (a) Show that, at the equator,the gravitational force on an object (the object's true weight) must exceed the object's apparent weight. (b) What are the apparent weights of a \(75.0-\mathrm{kg}\) person at the equator and at the poles? (Assume Earth is a uniform sphere and take \(g=9.800 \mathrm{~m} / \mathrm{s}^{2}\).)
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Physical Quantities And Units

Read Explanation

Oscillations

Read Explanation

Engineering Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.