/*! 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 29 A size-5 soccer ball of diameter... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A size-5 soccer ball of diameter 22.6 \(\mathrm{cm}\) and mass 426 \(\mathrm{g}\) rolls up a hill without slipping, reaching a maximum height of 5.00 \(\mathrm{m}\) above the base of the hill. We can model this ball as a thin- walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then?

Short Answer

Expert verified
(a) Angular velocity: 87.6 rad/s. (b) Rotational kinetic energy: 12.55 J.

Step by step solution

01

Understand the Problem

We have a soccer ball modeled as a thin-walled hollow sphere rolling up a hill and we need to determine its rotational rate at the base and its rotational kinetic energy at that point. We use energy conservation to find both answers.
02

Define the Known Values

Given a soccer ball with a diameter of 22.6 cm, mass of 426 g, and a maximum height of 5.00 m. Hence, the radius is half of the diameter, so the radius \( r = \frac{22.6}{2} = 11.3 \) cm or 0.113 m. Mass \( m = 426 \) g or 0.426 kg. Potential height \( h = 5.00 \) m.
03

Use the Conservation of Energy Principle

At the base of the hill, all energy is kinetic: translational and rotational. At the top, the energy is entirely potential. Thus, \( mgh = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 \). Here \( I \) is the moment of inertia, \( \omega \) is the angular velocity, and \( v \) is the linear speed at the base.
04

Determine the Moment of Inertia

The moment of inertia for a thin-walled hollow sphere is given by \( I = \frac{2}{3}mr^2 \). Substituting the known values: \( I = \frac{2}{3} \times 0.426 \times (0.113)^2 \), calculate \( I \approx 0.00328 \ \text{kg} \cdot \text{m}^2 \).
05

Solve for Linear Speed at the Base

Using the conservation of energy equation: \( 0.426 \times 9.8 \times 5.00 = \frac{1}{2} \times 0.426 \times v^2 + \frac{1}{2} \times 0.00328 \times \omega^2 \) and \( v = r\omega \). Substitute \( \omega \) in terms of \( v \) and solve, find \( v = \sqrt{2gh} = 9.9 \) m/s approximately, then \( \omega = \frac{v}{r} \).
06

Calculate Angular Velocity

Substitute the expression for \( \omega \) into the velocity equation: \( \omega = \frac{9.9}{0.113} \approx 87.6 \ \text{rad/s} \).
07

Calculate Rotational Kinetic Energy

The rotational kinetic energy is \( K = \frac{1}{2} I \omega^2 \). Calculate: \( K = \frac{1}{2} \times 0.00328 \times (87.6)^2 \) to find \( K \approx 12.55 \ \text{J} \).

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.

Conservation of Energy
The conservation of energy principle is crucial when dealing with rotational motion problems. It states that energy cannot be created or destroyed, only transformed from one form to another. In this exercise, we see energy transformation from kinetic to potential energy as the soccer ball rolls up the hill without slipping. At the bottom of the hill, all the energy is kinetic, comprising both translational and rotational kinetic energy. As the ball climbs the hill, this energy converts into gravitational potential energy. By setting the total energy at the base equal to the energy at the top, we can solve for unknown quantities like angular velocity.
  • Base state: Maximum kinetic energy (translational + rotational)
  • Top state: Maximum potential energy
This principle allows us to write the equation: \[mgh = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2\] where \(mgh\) is the potential energy, \(\frac{1}{2} mv^2\) is the translational kinetic energy, and \(\frac{1}{2} I\omega^2\) is the rotational kinetic energy.
Moment of Inertia
The moment of inertia (often denoted as \(I\)) is a property that quantifies how much an object resists angular acceleration around a particular axis. It plays a similar role in rotational motion as mass does in linear motion. For a thin-walled hollow sphere, the formula for the moment of inertia is \[I = \frac{2}{3}mr^2\] where \(m\) is the mass and \(r\) is the radius of the sphere. This formula reflects how the mass distribution affects rotational motion. In the context of our soccer ball, with a given radius and mass, we calculated \(I\) to be approximately 0.00328 kg⋅m². This value helps in determining the rotational kinetic energy and finding the angular velocity.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation. Like translational kinetic energy, it depends on the motion, but here it’s the angular motion. The formula to calculate rotational kinetic energy is:\[K = \frac{1}{2} I \omega^2\]where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For our soccer ball, using the known values of \(I\) and the calculated \(\omega\), we find that the rotational kinetic energy at the base of the hill is approximately 12.55 J. This energy contributes to the total energy the soccer ball carries as it starts to climb the hill.
Thin-Walled Hollow Sphere
A thin-walled hollow sphere, in physics, is an idealized model that simplifies the relationship between various properties like mass, radius, and angular motion. This model assumes the mass is distributed evenly along the edge of the sphere, with the inside being empty. For practical purposes, the soccer ball is modeled as such due to its structure and how it rolls without slipping.
  • Mass distribution affects moment of inertia
  • Simplifies calculations in rotational motion
By considering the ball as a thin-walled hollow sphere, we're able to easily apply the formula for the moment of inertia (\[I = \frac{2}{3}mr^2\]) and analyze the rotational aspects accurately. Understanding this model helps in simplifying complex problems and focuses on essential aspects like energy conservation and rotational dynamics.

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 uniform, \(0.0300-\mathrm{kg}\) rod of length 0.400 \(\mathrm{m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 \(\mathrm{kg}\) , are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 \(\mathrm{m}\) on each side of the center of the rod, and the system is rotating at 30.0 rev/min. With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?

The flywheel of an engine has moment of inertia 2.50 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 \(\mathrm{rev} / \mathrm{min}\) in 8.00 \(\mathrm{s}\) starting from rest?

A lawn roller in the form of a thin-walled, hollow cylinder with mass \(M\) is pulled horizontally with a constant horizontal force \(F\) applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force.

Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 \(\mathrm{cm}\) and a mass of 6.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end.

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?

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.