/*! 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 112 A physics major is working to pa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A physics major is working to pay his college tuition by perforning in a traveling camival. Herides a motorcycle inside a hollow transparent plastic sphere. After gaining sufficient speed, he travels in a vertical circle with a radius of 13.0 \(\mathrm{m}\) . The physics major has mass 70.0 \(\mathrm{kg}\) , and his motorcycle has mass 40.0 \(\mathrm{kg}\) . (a) What minimum speed must he have at the top of the circle if the tires of the motoreycle are not to lose contact with the sphere? (b) At the bottom of the circle, his speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

Short Answer

Expert verified
(a) Minimum speed at the top is approximately 11.3 m/s. (b) Normal force at the bottom is approximately 5409.3 N.

Step by step solution

01

Identify Forces at the Top of the Circle

At the top of the circle, the only two forces acting on the rider and motorcycle system are gravity and the normal force exerted by the sphere. For the system to maintain contact with the sphere, the normal force should be zero at minimum speed.
02

Apply Newton's Second Law

At the top, the gravitational force must equal the required centripetal force to keep moving in a circle. Use Newton's second law: \[ mg = \frac{mv^2}{r} \]where \( m \) is the combined mass \( m = 70 + 40 = 110\,\text{kg} \), and \( r = 13.0\,\text{m} \).
03

Solve for Minimum Speed at the Top

Rearrange the equation to solve for the speed \( v \):\[ v^2 = rg \]Substitute \( r = 13.0\,\text{m} \) and \( g = 9.81\,\text{m/s}^2 \): \[ v^2 = 13.0 \times 9.81 \]\[ v = \sqrt{13.0 \times 9.81} \approx 11.3 \text{ m/s} \].
04

Apply Concepts at the Bottom of the Circle

At the bottom, the speed of the motorcycle is twice the minimum speed calculated. Thus, \( v = 2 \times 11.3 = 22.6 \text{ m/s} \).
05

Compute Normal Force at the Bottom

Use the centripetal force formula at the bottom where the gravitational force and normal force contribute to centripetal acceleration:\[ N - mg = \frac{mv^2}{r} \]Rearrange to find \( N \):\[ N = mg + \frac{mv^2}{r} \].
06

Calculate the Normal Force

Substitute \( m = 110 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), \( v = 22.6 \text{ m/s} \), and \( r = 13.0 \text{ m} \) into the equation:\[ N = 110 \times 9.81 + \frac{110 \times (22.6)^2}{13.0} \]\[ N = 1079.1 + \frac{110 \times 511.76}{13} \]\[ N = 1079.1 + 4330.18 \approx 5409.3 \text{ N} \].

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.

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that relates the net force acting on an object to its acceleration. It's mathematically expressed as:
  • Force equals mass times acceleration (\[ F = ma \]).
In the context of the carnival rider, this law helps us analyze the forces acting when he's moving in a circle. At the top of the vertical circle, the gravitational force provides the necessary centripetal force for circular motion.
In simpler terms, gravity is responsible for "pulling" the rider and his motorcycle downwards in such a way that they keep moving in the circle's path, without falling off. For circular motion to be possible, the gravitational force needs to be perfectly balanced with the centripetal force, especially at the top of the circle.
This is represented by the equation:\[mg = \frac{mv^2}{r}\]where:
  • \(m\) is the total mass (rider + motorcycle).
  • \(g\) is gravitational acceleration.
  • \(v\) is the speed at the top of the circle.
  • \(r\) is the radius of the circle.
By setting these forces equal, we can determine the minimum speed needed at the top of the circle to maintain contact.
Normal Force
Normal force is the supportive force exerted by a surface to keep objects in contact. It's essential in situations involving surfaces, like our carnival rider inside the sphere.
In a vertical circular motion, normal force changes with position. At the top of the circle, minimal speed means the normal force is zero. Why zero? That's because at the least speed required not to fall, gravity alone is doing all the work.
It's like balancing at the brink—enough speed keeps the gravitational pull in check without the sphere having to "push back".
  • If speed increases, then normal force isn't zero anymore.
  • At the bottom of the circle, both the normal and gravitational forces provide necessary centripetal force.
To find the normal force at the circle's bottom, considering the higher speed, we use:\[N - mg = \frac{mv^2}{r}\]Rearranging gives:\[N = mg + \frac{mv^2}{r}\]At the bottom, speed doubles, which amplifies the centripetal force needed significantly. Thus, the normal force is greater compared to other positions in the circle.
Gravitational Force
Gravitational force is the attractive force between two objects with mass. On Earth, it gives weight to objects and causes them to fall when dropped.
For the physics major riding the motorcycle inside the sphere, gravity plays a pivotal role.
  • At the top of the sphere, gravitational force acts downwards, crucially contributing to the centripetal force needed to keep moving along the circle.
  • Together with speed, it ensures the rider doesn’t lose contact with the sphere surface.
The force can be calculated using:\[mg\] where \( m \) is the mass and \( g \) is the acceleration due to gravity. In circular motion, this force helps dictate the minimum speed required at different points of the circle.
By understanding how gravity affects motion in circular paths, one can ensure stability and balance, crucial for performances like those at carnivals or any context involving vertical loops.

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 small block with mass \(m\) rests on a frictionless horizontal tabletop a distance \(r\) from a hole in the center of the table (Fig. 5.79\() .\) A string tied to the small block passes down through the hole, and a larger block with mass \(M\) is suspended from the free end of the string. The small block is set into uniform circular motion with radius \(r\) and speed \(v\) . What must \(v\) be if the large block is to remain motionless when released? figure can't copy

A model airplane with mass 2.20 \(\mathrm{kg}\) moves in the \(x y\) -plane such that its \(x-\) and \(y\) -coordinates vary in time according to \(x(t)=\alpha-\beta t^{3}\) and \(y(t)=y t-\delta t^{2},\) where \(\alpha=1.50 \mathrm{m}, \beta=\) \(0.120 \mathrm{m} / \mathrm{s}^{3}, \gamma=3.00 \mathrm{m} / \mathrm{s},\) and \(\delta=1.00 \mathrm{m} / \mathrm{s}^{2} .\) (a) Calculate the \(x-\) and \(y\) -components of the net force on the plane as functions of time. (b) Sketch the trajectory of the airplane between \(t=0\) and \(t=3.00 \mathrm{s}\) , and draw on your sketch vectors showing the net force on the airplane at \(t=0, t=1.00 \mathrm{s}, t=2.00 \mathrm{s},\) and \(t=3.00 \mathrm{s}\) . For each of these times, relate the direction of the net force to the direction that the airplane is turning, and to whether the airplane is speeding up or slowing down (or neither). (c) What are the magnitude and direction of the net force at \(t=3.00 \mathrm{s} ?\)

Friction in an Elevator. You are riding in an elevator on the way to the 18th floor of your dormitory. The elevator is accelerating upward with \(a=190 \mathrm{m} / \mathrm{s}^{2}\) . 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 750.0-kg boulder is raised from a quarry 125 \(\mathrm{m}\) deep by a long uniform chain having a mass of 575 \(\mathrm{kg}\) . This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.50 times its weight without breaking. (a) What is the maximum acceleration the boulder can have and still get out of the quarry, and (b) how long does it take to be lifted out at maximum acceleration if it started from rest?

A rock with mass \(m=3.00 \mathrm{kg}\) falls from rest in a viscous medium. The rock is acted on by a net constant downward force of 18.0 \(\mathrm{N}\) (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force \(f=k v,\) where \(v\) is the speed in \(\mathrm{m} \mathrm{m} / \mathrm{s}\) and \(k=2.20 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}\) (see Section 5.3\() .\) (a) Find the initial acceleration \(a_{0} \cdot\) (b) Find the acceleration when the speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) (c) Find the speed when the acceleration equals 0.1\(a_{0}\) (d) Find the terminal speed \(v_{t}\) (e) Find the coordinate, speed, and acceleration 2.00 s after the start of the motion. (f) Find the time required to reach a speed 0.9\(v_{\mathrm{t}}\)

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.