/*! 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 45 A small remote-controlled car wi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A small remote-controlled car with mass \(1.60 \mathrm{~kg}\) moves at a constant speed of \(v=12.0 \mathrm{~m} / \mathrm{s}\) in a track formed by a vertical circle inside a hollow metal cylinder that has a radius of \(5.00 \mathrm{~m}\) (Fig. E5.45). What is the magnitude of the normal force exerted on the car by the walls of the cylinder at (a) point \(A\) (bottom of the track) and (b) point \(B\) (top of the track)?

Short Answer

Expert verified
The magnitudes of the normal force exerted on the car by the walls of the track at point A and point B can be calculated following the steps above. Make sure to properly plug in all given quantities in the correct equations.

Step by step solution

01

Writing down given quantities

First, list down all the given quantities. Mass of the car, \(m = 1.60 kg\), Speed of the car, \(v = 12.0 m/s\), Radius of the track, \(r = 5.00 m\), Acceleration due to gravity, \(g = 9.81 m/s^2\). Assuming, initially, up direction is positive.
02

Calculate the gravitational force acting on the car

Calculate the gravitational force acting on the car using the formula \(F_g = m*g\). Substituting the given values, \(F_g = 1.60 kg * 9.81 m/s^2 = 15.696 N\). Now the gravitational force is acting downwards.
03

Calculate Normal force at point A

At the bottom of the track (point A), the normal force and the gravitational force both act upwards. The sum of these forces provides the centripetal force needed for circular motion which can be given as \(F_c = m*v^2/r\). Following Newton's second law, the equation of motion at point A (bottom of the track) is, \(F_N + F_g = F_c\). Solving for the normal force, we get \(F_N = F_c - F_g\). After substituting the values into the equation, we can solve for \(F_N\), which is the magnitude of the normal force at point A.
04

Calculate Normal force at point B

At point B (top of the track), the gravitational force acts in the same direction as the centripetal force, while the normal force acts in the opposite direction. The centripetal force required for circular motion is given as \(F_c = m*v^2/r\). According to Newton's second law, the equation of motion at point B is \(F_N = F_c - F_g\). After plugging in the values, we can solve for \(F_N\), which is the magnitude of the normal force at point B.

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 91Ó°ÊÓ!

Key Concepts

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

Centripetal Force
Centripetal force is the invisible pull that keeps an object moving in a circular path; it's always directed towards the center of the circle. For objects in circular motion, like a car speeding around a track, centripetal force doesn't exist by itself; it's the result of other forces, such as tension, gravity, or in this case, the normal force exerted by the walls of a cylinder.

Think of it when you're swinging a ball attached to a string; your hand provides the force necessary to keep the ball in its circular path. Mathematically, this force can be described by the equation: \( F_c = m\frac{v^2}{r} \), where \( m \) is the mass of the object, \( v \) is its velocity, and \( r \) is the radius of the circle. Without sufficient centripetal force, an object will fly out of its circular path, following a straight line due to inertia.
Gravitational Force
Gravitational force is a universal force of attraction acting between all matter. According to Newton's law of universal gravitation, every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. On Earth, this force causes objects to fall towards the ground and is described by the simple equation \( F_g = m \times g \), where \( F_g \) is the gravitational force, \( m \) is the mass, and \( g \) is the acceleration due to gravity, which is approximately \(9.81 m/s^2\) near the Earth's surface.

In circular motion exercises, gravity can play a significant role, especially in vertical paths, as it constantly affects the centripetal force required to maintain motion.
Newton's Second Law of Motion
Newton's second law of motion is the foundation of classical mechanics and states that the acceleration of an object is directly proportional to the net force applied and inversely proportional to its mass. Formally, the law is expressed by the equation \( F = m \times a \), where \( F \) represents the force applied, \( m \) is the mass, and \( a \) is the acceleration.

This law is pivotal in explaining how forces affect the motion of objects. When the forces are unbalanced, the object will accelerate. In the case of circular motion, this acceleration is directed towards the center of the path and is responsible for changing the direction of velocity, not its magnitude, because of the constant speed.
Circular Motion Physics
Circular motion physics involves studying movements where objects travel in a circular path. This can be either uniform motion, where the speed is constant, or non-uniform, where it changes. The critical thing to remember is that even if the speed is constant, velocity is not because velocity also depends on direction. Therefore, an object in circular motion is always accelerating, even at constant speed, because its direction is continually changing.

In the context of the remote-controlled car problem, we see how circular motion principles allow us to calculate the normal force at different points along the track. Furthermore, understanding the balance of forces involved in circular motion helps us predict the car's movement and the forces it experiences.

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 stockroom worker pushes a box with mass \(16.8 \mathrm{~kg}\) on a horizontal surface with a constant speed of \(3.50 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between the box and the surface is \(0.20 .\) (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

You are sitting on the edge of a horizontal disk (for example, a playground merry-go-round) that has radius \(3.00 \mathrm{~m}\) and is rotating at a constant rate about a vertical axis. (a) If the coefficient of static friction between you and the surface of the disk is \(0.400,\) what is the minimum time for one revolution of the disk if you are not to slide off? (b) Your friend's weight is half yours. If the coefficient of static friction for him is the same as for you, what is the minimum time for one revolution if he is not to slide off?

If the person steps onto a smooth rock surface that's inclined at an angle large enough that these shoes begin to slip, what will happen? (a) She will slide a short distance and stop; (b) she will accelerate down the surface; (c) she will slide down the surface at constant speed; (d) we can't tell what will happen without knowing her mass.

Jack sits in the chair of a Ferris wheel that is rotating at a constant \(0.100 \mathrm{rev} / \mathrm{s}\). As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one- fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.

The cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of \(100 \mathrm{~m}\). Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every \(60.0 \mathrm{~s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs \(882 \mathrm{~N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

See all solutions

Recommended explanations on Physics Textbooks

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.