/*! 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 97 If the body's center of mass wer... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the body's center of mass were not placed on the rotational axis of the turntable, how would the person's measured moment of inertia compare to the moment of inertia for rotation about the center of mass? (a) The measured moment of inertia would be too large; (b) the measured moment of inertia would be too small; (c) the two moments of inertia would be the same; (d) it depends on where the body's center of mass is placed relative to the center of the turntable.

Short Answer

Expert verified
The correct answer is (a): The measured moment of inertia would be too large.

Step by step solution

01

Understanding the Parallel Axis Theorem

The Parallel Axis Theorem is a fundamental principle in classical mechanics. It states that the moment of inertia about any axis parallel to and a distance \(d\) away from an axis through the center of mass is the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the axes.
02

Apply the Parallel Axis Theorem to the Problem

In this case, the body's center of mass is not placed on the rotational axis of the turntable. Therefore, the moment of inertia is given by the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the center of mass and the rotational axis of the turntable.
03

Conclude the Answer based on the Theorem

According to the step 2, the measured moment of inertia when the center of mass is not on the rotation axis of the turntable will be greater than the moment of inertia about the center of mass. Therefore, the measured moment of inertia would be too large.

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.

Parallel Axis Theorem
The Parallel Axis Theorem is an essential tool in understanding rotational dynamics. This theorem helps us calculate the moment of inertia when an object's axis of rotation does not pass through its center of mass. Imagine an object rotating around an axis that is not passing through its center. The theorem tells us that the moment of inertia about this new axis can be found by taking the moment of inertia about the object's center of mass and adding to it the product of the object's mass and the square of the distance (\(d\)) between the two axes.Think of it like this:
  • We start with the object spinning perfectly around its center, where the inertia is known.
  • When we shift the axis, we imagine dragging the rotation outwards, which adds inertia due to increased distance.
The formula looks like this:\[ I = I_{cm} + md^2 \]where \( I \) is the moment of inertia about the new axis, \( I_{cm} \) is the inertia at the center of mass, \( m \) is the mass, and \( d \) is the distance between the axes.This relationship elegantly quantifies how much additional inertia results from moving the axis of rotation away from the center.
Rotational Motion
Rotational motion occurs when an object spins around an axis. This concept is similar to linear motion but involves entire bodies rotating in circles or paths. Understanding rotational motion requires grasping several core terms and ideas. Key Terms:
  • Axis of Rotation: This is the line around which an object rotates.
  • Angular Velocity: This indicates how fast an object is spinning.
  • Torque: The measure of force that causes the object to rotate.
A few important ideas govern rotational motion: - Every point on a rotating body moves in a circular path, centered on the axis of rotation. - The moment of inertia is crucial here as it measures how much torque is needed for a given angular acceleration. Greater moments mean more force is needed to achieve the same rotational speed. In essence, rotational motion is the angular counterpart to the more familiar linear motion. Think of a spinning wheel: its spokes, rim, and whole body move because of this circular motion around its central axis.
Center of Mass
The center of mass is a point where the entire mass of a body or system can be considered as concentrated. It’s like the balancing point of an object, where it remains balanced in all directions. To visualize the center of mass, imagine a uniform stick. If you balance it perfectly on your finger without it tipping to one side, you've found its center of mass. In more complex systems, such as irregular shaped bodies, the center of mass could be outside the physical material, like near the edge or an empty space inside. Importance in Physics:
  • It's crucial in calculating the motion of an object since forces and reactions are often applied at this point.
  • The center of mass allows simplification of real-world problems, focusing on a single point rather than complex shapes.
  • In rotational dynamics, it serves as the pivot point for analyzing torque and moment of inertia.
In any moving or rotating system, the center of mass is integral in predicting motion paths and behaviors. It’s not always where there's material but is always where the math makes it balance.

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 demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel \(0.650 \mathrm{~m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects \(0.200 \mathrm{~m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is \(8.00 \mathrm{~kg} ;\) its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at \(5.00 \mathrm{rev} / \mathrm{s}\). Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at \(0.300 \mathrm{rev} / \mathrm{s}\). (d) At what rate must the shaft rotate in order that it may be supported at one end only?

A large uniform horizontal turntable rotates freely about a vertical axle at its center. You measure the radius of the turntable to be \(3.00 \mathrm{~m} .\) To determine the moment of inertia \(I\) of the turntable about the axle, you start the turntable rotating with angular speed \(\omega\), which you measure. You then drop a small object of mass \(m\) onto the rim of the turntable. After the object has come to rest relative to the turntable, you measure the angular speed \(\omega_{\mathrm{f}}\) of the rotating turntable. You plot the quantity \(\left(\omega-\omega_{\mathrm{f}}\right) / \omega_{\mathrm{f}}\) (with both \(\omega\) and \(\omega_{\mathrm{f}}\) in rad \(\left./ \mathrm{s}\right)\) as a function of \(m\) (in kg). You find that your data lie close to a straight line that has slope \(0.250 \mathrm{~kg}^{-1}\). What is the moment of inertia \(I\) of the turntable?

Two uniform solid balls are rolling without slipping at a constant speed. Ball 1 has twice the diameter, half the mass, and one-third the speed of ball 2 . The kinetic energy of ball 2 is \(27.0 \mathrm{~J}\). What is the kinetic energy of ball \(1 ?\)

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3 ). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\). If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{\mathrm{k}}\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0\) Rolling without slipping sets in when \(v_{x}=r \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

A \(12.0 \mathrm{~kg}\) box resting on a horizontal, frictionless surface is attached to a \(5.00 \mathrm{~kg}\) weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass \(2.00 \mathrm{~kg}\) and diameter \(0.500 \mathrm{~m} .\) After the system is released, find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force that the axle exerts on the pulley.

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.