/*! 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 The parallel axis theorem provid... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+M h^{2},\) where \(I_{\mathrm{cm}}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

Short Answer

Expert verified
The moment of inertia is \(I = \frac{3}{2}MR^2\).

Step by step solution

01

Identify Parameters

We need the moment of inertia (\(I\)) about an axis that lies on the surface of the cylinder and is perpendicular to the circular ends. We know the parallel axis theorem formula: \(I=I_{\mathrm{cm}}+Mh^2\). We need \(I_{\mathrm{cm}}\), which is the moment of inertia of a cylinder about its central axis (perpendicular to its length). For a solid cylinder, \(I_{\mathrm{cm}} = \frac{1}{2}MR^2\), where \(M\) is the mass of the cylinder and \(R\) is its radius.
02

Determine Perpendicular Distance \(h\)

The perpendicular distance \(h\) between the central axis and the surface axis is the radius of the cylinder, \(R\). This is because the axis of interest is offset from the central axis by exactly the radius of the cylinder.
03

Apply the Parallel Axis Theorem

Substitute \(I_{\mathrm{cm}}\) and \(h\) into the parallel axis theorem equation: \[I = I_{\mathrm{cm}} + Mh^2 = \frac{1}{2}MR^2 + MR^2.\] This simplifies to \[I = \frac{3}{2}MR^2.\]
04

Conclusion

The expression for the moment of inertia of the solid cylinder about the specified axis is \(I = \frac{3}{2}MR^2\).

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.

Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics. It's often compared to mass in linear motion because it measures an object's resistance to rotational acceleration about an axis. The larger the moment of inertia, the harder it is to change the object's rotation. For any given object or shape, the moment of inertia depends on both the mass distribution and the axis of rotation.

A helpful way to understand this is to imagine yourself holding a spinning object. If the mass is spread out, it's more difficult to stop or start spinning it. This distribution of mass is expressed as the moment of inertia.

In calculations, the moment of inertia is noted with the symbol \(I\). The mathematical expression varies depending on the shape and orientation. For example, for a solid cylinder rotating around its central axis, the moment of inertia \(I_{\mathrm{cm}} = \frac{1}{2}MR^2\). Here, \(M\) represents mass and \(R\) represents the radius.
Solid Cylinder
Solid cylinders are common geometric shapes analyzed in physics problems, especially those involving rotation. A solid cylinder has an even mass distribution along its entire shape, making it straightforward to calculate properties like the moment of inertia.

When discussing a solid cylinder in terms of moment of inertia, it's crucial to consider the axis about which it rotates. A cylinder can rotate around its central axis (the vertical line through its height) or any other axis, like one on its surface.

Because the mass of a solid cylinder is symmetrically distributed, calculations for properties like moment of inertia can often be simplified. For the central axis, for instance, the formula is \(I_{\mathrm{cm}} = \frac{1}{2}MR^2\). This simplicity is due to the fact that the distance from each infinitesimal mass element to the axis is uniformly distributed, allowing us to apply standard results.
Perpendicular Axis
When working with the parallel axis theorem, understanding the concept of perpendicular axes is vitally important. In the context of the problem about the solid cylinder, consider the primary axis of interest. This axis lies on the surface of the cylinder and is perpendicular to its circular ends.

The perpendicular distance from the central axis (which passes through the center of mass and is often the axis of easiest calculation) to this new axis on the surface is defined as \(h\). For a solid cylinder, this distance \(h\) is equal to the cylinder's radius \(R\).

It's noteworthy because moving the axis from the center out to the surface increases the moment of inertia. The application of the parallel axis theorem helps us understand and calculate such changes in the rotational dynamics of the object.
Center of Mass
The center of mass is a critical concept not just for motion but specifically within rotational dynamics, as it serves as a pivotal axis for calculations like those involving the moment of inertia. The center of mass is essentially the average position of all the mass in the object. It's where you can imagine the entire mass of an object being concentrated.

In regular, symmetrical objects, such as a solid cylinder, the center of mass is at its geometric center. This point acts as a natural rotational axis because any rotation around it does not affect the distribution of mass distances significantly, allowing for simpler moment of inertia calculations.

Understanding the center of mass is indispensable when using the parallel axis theorem. The theorem itself uses the inertia about this center point \( I_{\mathrm{cm}} \) to find inertia about any parallel axis. This helps in calculating rotational properties for different applied forces or rotational paths.

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

Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

A platform is rotating at an angular speed of \(2.2 \mathrm{rad} / \mathrm{s}\). A block is resting on this platform at a distance of \(0.30 \mathrm{~m}\) from the axis. The coefficient of static friction between the block and the platform is \(0.75\). Without any external torque acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform and determine the smallest distance from the axis at which the block can be relocated and still remain in place as the platform rotates.

A baggage carousel at an airport is rotating with an angular speed of \(0.20 \mathrm{rad} / \mathrm{s}\) when the baggage begins to be loaded onto it. The moment of inertia of the carousel is 1500 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Ten pieces of baggage with an average mass of \(15 \mathrm{~kg}\) each are dropped vertically onto the carousel and come to rest at a perpendicular distance of \(2.0 \mathrm{~m}\) from the axis of rotation. (a) Assuming that no net external torque acts on the system of carousel and baggage, find the final angular speed. (b) In reality, the angular speed of a baggage carousel does not change. Therefore, what can you say qualitatively about the net external torque acting on the system?

Concept Questions The drawing shows two identical systems of objects; each consists of three small balls (masses \(m_{1}, m_{2}\), and \(m_{3}\) ) connected by massless rods. In both systems the axis is perpendicular to the page, but it is located at a different place, as shown. (a) Do the systems necessarily have the same moments of inertia? If not, why not? (b) The same force of magnitude \(F\) is applied to the same ball in each system (see the drawing). Is the magnitude of the torque created by the applied force greater for system A or for system B? Or is the magnitude the same in the two cases? Explain. (c) The two systems start from rest. Will system A or system B have the greater angular speed at the same later time? Or will they have the same angular speeds? Justify your answer. Problem The masses of the balls are \(m_{1}=9.00 \mathrm{~kg}, m_{2}=6.00 \mathrm{~kg}\), and \(m_{3}=7.00 \mathrm{~kg}\). The magnitude of the force is \(F=424 \mathrm{~N}\). (a) For each of the two systems, determine the moment of inertia about the given axis of rotation. (b) Calculate the torque (magnitude and direction) acting on each system. (c) Both systems start from rest, and the direction of the force moves with the system and always points along the \(4.00-\mathrm{m}\) rod. What is the angular velocity of each system

In outer space two identical space modules are joined together by a massless cable. These modules are rotating about their center of mass, which is at the center of the cable, because the modules are identical (see the drawing). In each module, the cable is connected to a motor, so that the modules can pull each other together. The initial tangential speed of each module is \(v_{0}-17 \mathrm{~m} / \mathrm{s}\). Then they pull together until the distance between them is reduced by a factor of two. Determine the final tangential speed \(v_{\mathrm{f}}\) for each module.

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.