/*! 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 41 A solid cylinder with a radius o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A solid cylinder with a radius of \(4.0 \mathrm{cm}\) has the same mass as a solid sphere of radius \(R\). If the cylinder and sphere have the same moment of inertia about their centers, what is the sphere's radius?

Short Answer

Expert verified
The radius of the sphere is \(R = 5\) cm

Step by step solution

01

Write the Formula for Cylinder and Sphere's Moment of Inertia

Write down the moments of inertia for a cylinder and sphere: \(I_{cylinder} = 0.5 \cdot m \cdot r^2\) and \(I_{sphere} = 0.4 \cdot m \cdot R^2\)
02

equate the moments of inertia

Since the solid cylinder and sphere have the same moment of inertia, we can equate the two formulas: \(0.5 \cdot m \cdot r^2 = 0.4 \cdot m \cdot R^2\)
03

Cancel out the Mass

Cancellation of 'm' leaves us with: \(0.5 \cdot r^2 = 0.4 \cdot R^2\)
04

Solve for R

Rearrange the equation and solve for R in terms of r: \(R = \sqrt{\(0.5/0.4\) \cdot r^2}\)
05

Substitute the Radius of the Cylinder

Substitute the given radius of the cylinder (r) into the equation to find the value of R: \(R = \sqrt{\(0.5/0.4\) \cdot (4)^2}\)
06

Calculation

Calculate the value to find the sphere's radius

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.

Solid Cylinder
Understanding the moment of inertia for different bodies is essential in physics, particularly when studying rotational motion. A solid cylinder is a common object encountered in this context. It's crucial to comprehend how its mass distribution affects its rotational properties.

For a solid cylinder rotating about its central axis, the moment of inertia can be calculated using the formula \( I_{cylinder} = 0.5 \cdot m \cdot r^2 \), where \(m\) is the mass and \(r\) the radius of the cylinder. This equation reveals that the moment of inertia depends not only on the mass but also on the square of the radius. Therefore, as the radius increases, the cylinder's resistance to changes in rotational speed increases quadratically. In the given problem, the radius is \(4.0 \mathrm{cm}\), which provides tangible information to calculate the exact rotational inertia for the cylinder.
Solid Sphere
Comparatively, a solid sphere exhibits different distribution of its mass, influencing its motion when rotating about an axis through its center. The formula for a sphere's moment of inertia is \( I_{sphere} = 0.4 \cdot m \cdot R^2 \), where \(R\) is the sphere's radius.

It is noteworthy that for a sphere, a slightly different constant is used (0.4 instead of 0.5 for the cylinder). This reflects the mass being more evenly distributed across the volume of the sphere, making it less resistant to rotational change compared to a solid cylinder with the same mass and radius. This is particularly important when solving physics problems involving rotational dynamics, where different shapes must be handled distinctly due to their unique geometrical properties.
Radius of Gyration
A related concept to the moment of inertia is the radius of gyration. It's defined as the hypothetical distance from the axis of rotation at which the entire mass of a body could be concentrated without changing its rotational inertia. Symbolically, it's expressed as \( k \), satisfying \( I = m \cdot k^2 \).

The radius of gyration provides an insightful way to compare how mass is distributed in different bodies, even if they share a similar moment of inertia. It encapsulates in a single value the effect of an object's shape and size on its rotational characteristics. For both the cylinder and sphere mentioned earlier, one could compute their respective radii of gyration to get a deeper understanding of their mass distribution relative to rotational motion.
Physics Problem Solving
Approaching physics problem solving often involves a step-by-step method. This systematic approach breaks complex problems into manageable pieces, helping students and researchers alike to tackle intricate scenarios effectively.

Starting with identifying the known quantities and determining the target variable is a fundamental first step. Next, selecting the appropriate physical laws or formulas that relate these quantities is key. In our example, we use the formulas for moments of inertia of both a solid cylinder and a solid sphere to set up an equation based on their equivalency. Manipulating the equation algebraically and substituting numerical values then leads to the solution. The notion is to transform abstract physics concepts into solvable mathematical exercises, thereby providing clarity and understanding through calculation.

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

The earth's radius is \(6.37 \times 10^{\circ} \mathrm{m} ;\) it rotates once every 24 hours. a. What is the earth's angular speed? b. Viewed from a point above the north pole, is the angular velocity positive or negative? c. What is the speed of a point on the equator? d. What is the speed of a point on the earth's surface halfway between the equator and the pole? (Hint: What is the radius of the circle in which the point moves?)

A trap-jaw ant has mandibles that can snap shut with some force, as you might expect from its name. The formidable snap is good for more than capturing prey. When an ant snaps its jaws against the ground, the resulting force can launch the ant into the air. Here are typical data: An ant rotates its mandible, of length \(1.30 \mathrm{mm}\) and \(\mathrm{mass} 130 \mu \mathrm{g}\) (which we can model as a uniform rod rotated about its end), at a high angular speed. As the tip strikes the ground, it undergoes an angular acceleration of \(3.5 \times 10^{8} \mathrm{rad} / \mathrm{s}^{2}\). If we assume that the tip of the mandible hits perpendicular to the ground, what is the force on the tip? How does this compare to the weight of a \(12 \mathrm{mg}\) ant?

To throw a discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about \(1.8 \mathrm{m}\). If the thrower takes \(1.0 \mathrm{s}\) to complete one revolution, starting from rest, what will be the speed of the discus at release?

The \(1.00-\mathrm{cm}-\) long second hand on a watch rotates smoothly. a. What is its angular velocity? b. What is the speed of the tip of the hand?

A child on a merry-go-round takes \(3.0 \mathrm{s}\) to go around once. What is his angular displacement during a 1.0 s time interval?

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.