/*! 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 39 A regulation table tennis ball i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A regulation table tennis ball is a thin spherical shell \(40 \mathrm{mm}\) in diameter with a mass of 2.7 g. What is its moment of inertia about an axis that passes through its center?

Short Answer

Expert verified
The moment of inertia of the table tennis ball is approximately \(7.2 × 10^{-7} kg.m^2\).

Step by step solution

01

Convert mass and diameter to standard units

The exercise gives us mass and diameter in g and mm respectively. First, convert them to standard units used in Physics for calculations. The mass \( m = 2.7 g = 0.0027 kg \) and the diameter \( D = 40 mm = 0.04 m \).
02

Calculate the radius

The radius \( r \) of the ball is half its diameter. Hence, \( r = \frac{D}{2} = \frac{0.04}{2} = 0.02 m \).
03

Calculate the moment of inertia

Substitute the mass \(m\) and radius \(r\) into the moment of inertia formula for a thin spherical shell: \(I = \frac{2}{3} m r^2\). This will give us the moment of inertia \(I = \frac{2}{3} * 0.0027 kg * (0.02 m)^2\).

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.

Physics Calculations
Physics is fundamentally about understanding the behavior of the universe through mathematics. Physics calculations are the backbone of nearly all scientific and engineering disciplines, enabling us to quantify observations and predict the outcomes of experiments and everyday events.

In the context of our exercise, we are exploring the moment of inertia, a key concept in rotational dynamics that describes an object's resistance to changes in its rotation. The calculation involves the mass distribution of the object and the axis about which it spins. To solve for the moment of inertia of a table tennis ball, appropriate units are essential. As laid out in the solution, we must use the International System of Units (SI units), converting grams to kilograms and millimeters to meters before applying the mathematical formula. Accuracy in these basic calculations is crucial, as any inconsistency can greatly influence the outcome and understanding of an object’s physical behavior.
Spherical Shell Inertia
When we talk about spherical shell inertia, we are dealing with a specific type of moment of inertia pertinent to hollow objects whose mass is distributed over the surface of a sphere. A spherical shell is a three-dimensional object, but unlike a solid sphere, its mass is not distributed throughout its volume.

The moment of inertia for a spherical shell is derived from a unique formula distinct from that used for a solid sphere. For a spherical shell, the moment of inertia around an axis through the center is given by the formula: \[I = \frac{2}{3} m r^2\], where 'm' is the shell's mass and 'r' is the radius of the sphere. This calculation assumes a uniform density and thickness of the shell, which simplifies the integration process necessary to come up with the formula. Understanding these nuances is fundamental for physics students looking to grasp the complexities of rotational motion and the forces that govern it.
Table Tennis Ball Physics
Delving into table tennis ball physics can offer fascinating insights into the principles of motion and dynamics. A regulation table tennis ball, as in our exercise, exemplifies a thin spherical shell, which is crucial when calculating its moment of inertia.

The physical properties of table tennis balls - lightweight, bouncy, and having a small radius - contribute to their unique behavior during play. Their inertial characteristics govern how they spin, wobble, and react upon hitting the racket or table. When solving physics problems involving such objects, it's essential to remember the practical applications: these calculations can help us understand energy conservation during a ball's flight and impact, and even the spin imparted by skilled players. By correctly computing the moment of inertia of a table tennis ball, students not only solve a theoretical exercise but also gain a deeper appreciation for the physical principles observed in sports and recreation.

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 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?

U.S. nickels have a mass of \(5.00 \mathrm{g}\) and are \(1.95 \mathrm{mm}\) thick. If you stack 3 nickels on a table, how far above the table is their center of gravity?

The bunchberry flower has the fastest-moving parts ever seen in a plant. Initially, the stamens are held by the petals in a bent position, storing energy like a coiled spring. As the petals release, the tips of the stamens fly up and quickly release a burst of pollen. Figure \(\mathrm{P} 7.78\) shows the details of the motion. The tips of the stamens act like a catapult, flipping through a \(60^{\circ}\) angle; the times on the earlier photos show that this happens in just 0.30 ms. We can model a stamen tip as a 1.0 -mm- long, \(10 \mu\) g rigid rod with a \(10 \mu g\) anther sac at one end and a pivot point at the opposite end. Though an oversimplification, we will model the motion by assuming the angular acceleration is constant throughout the motion. How large is the "straightening torque"? (You can omit gravitational forces from your calculation; the gravitational torque is much less than this.) A. \(2.3 \times 10^{-7} \mathrm{N} \cdot \mathrm{m}\) B. \(3.1 \times 10^{-7} \mathrm{N} \cdot \mathrm{m}\) C. \(2.3 \times 10^{-5} \mathrm{N} \cdot \mathrm{m}\) D. \(3.1 \times 10^{-5} \mathrm{N} \cdot \mathrm{m}\)

An athlete at the gym holds a \(3.0 \mathrm{kg}\) steel ball in his hand. His arm is \(70 \mathrm{cm}\) long and has a mass of \(4.0 \mathrm{kg}\). What is the magnitude of the gravitational torque about his shoulder if he holds his arm a. Straight out to his side, parallel to the floor? b. Straight, but \(45^{\circ}\) below horizontal?

A car with 58-cm-diameter tires accelerates uniformly from rest to \(20 \mathrm{m} / \mathrm{s}\) in \(10 \mathrm{s}\). How many times does each tire rotate?

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.