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A uniform sphere made of modeling clay has radius \(R\) and moment of inertia \(I_{1}\) for rotation about a diameter. It is flattened to a disk with the same radius \(R .\) In terms of \(I_{1},\) what is the moment of inertia of the disk for rotation about an axis that is at the center of the disk and perpendicular to its flat surface?

Short Answer

Expert verified
The moment of inertia of the disk is \(\frac{5}{4}\) times the moment of inertia of the sphere.

Step by step solution

01

Remember Moment of Inertia Formulas

The moment of inertia of a body depends not only on its mass distribution but also on the location of the axis of rotation. For a solid sphere rotating about an axis passing through its center, the moment of inertia I is given by \(\frac{2}{5}MR^2\), where M is the mass and R is the radius of the sphere. On the other hand, for a thin circular disc, or a cylindrical shell rotating about an axis perpendicular to the plane of the disk and passing through its center, the moment of inertia I is given by \(\frac{1}{2}MR^2\). Since the mass and radius of the clay didn't change after it was flattened, the mass and the radius of the sphere and the disk are the same.
02

Relate The Two Moments

In terms of \(I_1\), we have \(I_1 = \frac{2}{5}MR^2\). Now write the moment of inertia of the disk \(I_2\) in terms of \(I_1\): Since the moment of inertia of a flat disk is \(I_2 = \frac{1}{2}MR^2\) and \(I_1 = \frac{2}{5}MR^2\), rearrange this equation to get: \(I_2 = \frac{5}{4}I_1\).
03

Interpretation

The moment of inertia of the disk is \(\frac{5}{4}\) of the moment of inertia of the original sphere. The shaping of the clay does not affect mass but spreads it out from the axis of rotation, which increases the moment of inertia.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Rotation Axis
When we talk about the moment of inertia, one key factor to consider is the location of the rotation axis. This axis is an imaginary line around which an object spins or rotates. It can be compared to a pivot point that determines the manner in which the body's parts contribute to the moment of inertia.
The distribution of mass relative to this axis greatly impacts how much effort is required to rotate the object. In the problem you provided, the axis of rotation is crucial because it changes when the shape of the object changes—from a sphere to a disk. For a sphere, the axis passes through its center; while for a disk, the axis is perpendicular to and passes through the center of its flat surface.
The change in the shape of the object and the associated change in the rotation axis are what lead to different moments of inertia. This change shows how intricately the axis and geometry of objects govern rotational dynamics.
Exploring the Solid Sphere
A solid sphere is a three-dimensional round object, like a ball. Its moment of inertia depends on its size and mass distribution. In mathematical terms, for a solid sphere, the moment of inertia about an axis through its center is calculated using the formula: \(\frac{2}{5}MR^2\). Here, \(M\) represents the mass of the sphere, and \(R\) is its radius.
This formula tells us that a significant portion of a sphere's mass is distributed around its center. The smaller the radius, the easier it is to rotate the sphere around this axis due to the alignment of mass towards the center. This centralized mass distribution plays a big role in defining the sphere's moment of inertia. It's why we calculate \(I_1\) using this specific formula in the provided exercise.
Understanding the Circular Disk
A circular disk can be visualized like a flat plate or a Frisbee, and it is this shape that the sphere in the exercise is flattened into, keeping the same radius. When calculating the moment of inertia for such a disk, we use the formula: \(\frac{1}{2}MR^2\). In this formula, \(M\) is the mass of the disk, and \(R\) is the radius.
The disk's mass distribution around its central axis significantly affects its rotational properties. Unlike a sphere, a disk has much of its mass spread out from the center, making the effort needed to rotate it different. This new distribution helps explain why the moment of inertia of the disk, \(I_2\), results in a different value compared to the sphere, specifically \(\frac{5}{4}I_1\), as calculated using the problem's solution.
Role of Mass Distribution in Rotational Inertia
Mass distribution refers to how mass is spread over the volume of an object. This factor is crucial in determining the moment of inertia. The more the mass is spread away from the axis of rotation, the larger the moment of inertia will be. This principle explains why different shapes with the same mass and size can have vastly different rotational behaviors.
In the exercise, when the spherical clay is flattened into a disk, its mass distribution changes. Although the mass stays constant, the shape and how that mass is spread out alters the inertia characteristics. The disk has its mass more evenly spread in a flat plane, which increases its moment of inertia compared to when it was a sphere. Understanding this link between shape, mass distribution, and inertia helps explain the nature of rotational dynamics in different objects—providing a solid foundation for understanding physics principles in rotational motion.

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Most popular questions from this chapter

The earth is approximately spherical, with a diameter of \(1.27 \times 10^{7} \mathrm{~m} .\) It takes 24.0 hours for the earth to complete one revolution. What are the tangential speed and radial acceleration of a point on the surface of the earth, at the equator?

Energy is to be stored in a \(70.0 \mathrm{~kg}\) flywheel in the shape of a uniform solid disk with radius \(R=1.20 \mathrm{~m}\). To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is \(3500 \mathrm{~m} / \mathrm{s}^{2}\). What is the maximum kinetic energy that can be stored in the flywheel?

A pulley on a frictionless axle has the shape of a uniform solid disk of mass \(2.50 \mathrm{~kg}\) and radius \(20.0 \mathrm{~cm}\). A \(1.50 \mathrm{~kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has \(4.50 \mathrm{~J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Measuring \(I\). As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be \(0.640 \mathrm{~m}\). Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an \(8.20 \mathrm{~kg}\) block of wood from the free end of the rope, as in Fig. E9.49. You release the system from rest and find that the block descends \(12.0 \mathrm{~m}\) in \(4.00 \mathrm{~s}\). What is the moment of inertia of the wheel for this axis?

A uniform wheel in the shape of a solid disk is mounted on a frictionless axle at its center. The wheel has mass \(5.00 \mathrm{~kg}\) and radius \(0.800 \mathrm{~m} .\) A thin rope is wrapped around the wheel, and a block is suspended from the free end of the rope. The system is released from rest and the block moves downward. What is the mass of the block if the wheel turns through 8.00 revolutions in the first \(5.00 \mathrm{~s}\) after the block is released?

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