/*! 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 25 A \(392-\mathrm{N}\) wheel comes... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(392-\mathrm{N}\) wheel comes off a moving truck and rolls with- out slipping along a highway. At the bottom of a hill it is rotating at 25.0 \(\mathrm{rad} / \mathrm{s} .\) The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2} .\) Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 \(\mathrm{J} .\) Calculate \(h .\)

Short Answer

Expert verified
The height \( h \) is approximately 11.73 meters.

Step by step solution

01

Understanding the problem

We need to calculate the height \( h \) that the wheel reaches as it rolls up a hill and comes to a stop. The work done by friction is 3500 J, and the kinetic energy initially present in the wheel involves both translational and rotational motion.
02

Calculate mass of the wheel

The weight of the wheel is given as 392 N. Using the relation \( W = mg \), where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, we calculate the mass \( M \) of the wheel:\[ M = \frac{392}{9.8} = 40 \, \text{kg} \]
03

Calculate initial kinetic energy

The initial kinetic energy \( K_i \) of the wheel comprises both translational and rotational components. The translational kinetic energy is given by \( \frac{1}{2} M v^2 \) and the rotational kinetic energy is \( \frac{1}{2} I \omega^2 \). First, we find the linear velocity \( v \) using \( v = R \omega \): \[ v = 0.6 \, \text{m} \times 25.0 \, \text{rad/s} = 15.0 \, \text{m/s} \]Then, calculate \( I = 0.800 MR^2 \),\[ I = 0.800 \times 40 \times (0.6)^2 = 11.52 \, \text{kg} \, \text{m}^2 \]Now, calculate both components:\[ K_{\text{trans}} = \frac{1}{2} \times 40 \times (15)^2 = 4500 \, \text{J} \]\[ K_{\text{rot}} = \frac{1}{2} \times 11.52 \times (25)^2 = 3600 \, \text{J} \]
04

Total initial kinetic energy

Add the translational and rotational kinetic energies to get the total initial kinetic energy:\[ K_i = 4500 \, \text{J} + 3600 \, \text{J} = 8100 \, \text{J} \]
05

Use conservation of energy

As the wheel reaches a height \( h \), its kinetic energy is converted into gravitational potential energy and the work done by friction (3500 J) reduces the total energy. Applying the conservation of energy:\[ 8100 = Mgh + 3500 \]Substitute \( M = 40 \, \text{kg} \) to find \( h \): \[ 8100 = 40 \times 9.8 \times h + 3500 \]
06

Solve for height h

Rearrange and solve for \( h \):\[ 4600 = 392h \]\[ h = \frac{4600}{392} \approx 11.73 \, \text{meters} \]

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.

Rotational Kinetics
Rotational kinetics plays a crucial role in understanding how objects like wheels behave when they rotate. Unlike linear motion, where we only deal with variables like speed and distance, rotational motion adds layers like angular velocity and moment of inertia. The angular velocity (\( \omega \)) is analogous to linear velocity but in terms of rotation around an axis. It measures how fast the object is spinning.
  • Angular Velocity (\(\omega\)): This is the rate of change of angular position of the wheel and is measured in radians per second.
  • Moment of Inertia (\(I\)): This is a property of the wheel that depends on its mass distribution; it describes how hard it is to get the wheel spinning or to stop it once it’s spinning.
In this problem, the rotational kinetic energy involves the moment of inertia and angular velocity. The formula for rotational kinetic energy is given by:\[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \]So, when dealing with wheels rolling uphill, both the rotational and translational kinetic energies must be considered to understand the overall energy dynamics.
Conservation of Energy
The principle of conservation of energy is one of the key ideas in physics. It states that energy cannot be created or destroyed, only transformed from one form to another. In this context, the kinetic energy of the wheel at the bottom of the hill is transformed into gravitational potential energy at the top, along with some of it being lost due to work done by friction.
Energy conservation can be neatly summarized as:\[ K_{\text{initial}} = K_{\text{final}} + W_{\text{friction}} \]
  • Kinetic Energy: Begins as both rotational and translational at the bottom.
  • Gravitational Potential Energy: Increases as the wheel goes up the hill, countering the initial kinetic energy.
  • Work Done by Friction: Represents energy lost due to friction, which does negative work on the system.
This principle guides us to calculate the height by equating the loss of kinetic energy to the gain in potential energy plus the energy lost to friction.
Friction Work
Friction work is an important factor whenever there is motion against a surface. It is the work done by the frictional force, which in most cases, works to slow down or stop moving objects. In this exercise, friction does 3500 J of work opposing the motion of the wheel as it climbs uphill.
  • Friction Work (\(W_f\)): Calculated as the product of friction force and the distance over which it acts.
  • Negative Work: Called 'negative' because it removes energy from the system rather than adds.
Friction is not just a nuisance; it plays a critical role in allowing the wheel to roll instead of slip, but it also doubles as an energy dissipation pathway. In problems involving conservation of energy, friction’s role is crucial in accurately determining how much energy is converted into heat or sound, reducing the energy available for work or motion.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy stored in an object due to its position in a gravitational field. The higher up the object is, the more gravitational potential energy it possesses. When the wheel rolls uphill, it gains GPE which is given by:\[ U = mgh \]where:
  • \(m\): Mass of the object (in this case, the wheel).
  • \(g\): Acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\) on Earth.
  • \(h\): Height above the starting point.
As the wheel ascends, its kinetic energy is continuously converted into gravitational potential energy. Understanding this conversion is key to determining the height at which the wheel will stop. In this exercise, we used the initial kinetic energy and the work done by friction to solve for the resulting height \(h\), illustrating how these types of energy interact and balance each other out.

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

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly \(10^{14}\) times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was \(7.0 \times 10^{5} \mathrm{km}\) (comparable to our sun); its final radius is 16 \(\mathrm{km} .\) If the original star rotated once in 30 days, find the angular speed of the neutron star.

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) She then tucks into a small ball, decreasing this moment of inertia to 3.6 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) While tucked, she makes two complete revolutions in 1.0 \(\mathrm{s}\) s. If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of 7.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; (b) the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

An engine delivers 175 hp to an aircraft propeller at 2400 rev/min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 \(\mathrm{N}\) is applied to the rim of the wheel. The wheel has radius 0.120 \(\mathrm{m} .\) Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

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.