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Chapter 9: Problem 70

A uniform disk has radius \(R_{0}\) and mass \(M_{0}\). Its moment of inertia for an axis perpendicular to the plane of the disk at the disk's center is \(\frac{1}{2} M_{0} R_{0}^{2}\). You have been asked to halve the disk's moment of inertia by cutting out a circular piece at the center of the disk. In terms of \(R_{0}\), what should be the radius of the circular piece that you remove?

Short Answer

Expert verified
The radius of the circular piece that you need to remove from the center of the disc in order to halve its moment of inertia is \(r = R_{0}\sqrt[4]{1/2}\)

Step by step solution

01

Define the Initial Moment of Inertia

The initial moment of inertia \(I_{0}\) of the uniform disk is given by \(I_{0} = \frac{1}{2}M_{0}R_{0}^{2}\)
02

Define the Final Moment of Inertia

The final moment of inertia \(I_{f}\) of the disk, after removing a smaller disc at the center, will be half of the initial moment of inertia, i.e., \(I_{f} = \frac{1}{2} I_{0} = \frac{1}{4}M_{0}R_{0}^{2}\)
03

Setup the Equation

The moment of inertia of the remaining disc is the difference between the original moment of inertia and the moment of inertia of the portion removed. If the radius of the portion removed is \(r\), its moment of inertia will be \(\frac{1}{2} m r^{2}\), where \(m\) is its mass. Since we are dealing with a uniform disc, \(m = M_{0} \times \frac{r^{2}}{R_{0}^{2}}\). Therefore, the moment of inertia of the removed portion is \(\frac{1}{2} M_{0} \frac{r^{2}}{R_{0}^{2}} r^{2} = \frac{1}{2}M_{0} \frac{r^{4}}{R_{0}^{2}}\). The equation to be solved is therefore, \( \frac{1}{2}M_{0}R_{0}^{2} - \frac{1}{2}M_{0} \frac{r^{4}}{R_{0}^{2}} = \frac{1}{4}M_{0}R_{0}^{2}\)
04

Solve the Equation

Simplifying the above equation, you obtain a relation between \( r\) and \(R_{0}\). The result is \(r = R_{0}\sqrt[4]{1/2}\).

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Key Concepts

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

Uniform Disk
A uniform disk is a circular, flat object where the mass is evenly distributed throughout its entire volume. This uniform distribution means that each small section of the disk has the same mass per unit area.

Understanding a uniform disk is crucial in physics when calculating quantities like the moment of inertia. The moment of inertia is a measure of an object's resistance to changes in its rotation. For a uniform disk, this rotational inertia depends solely on the disk's mass and its radius.

To find the moment of inertia of a uniform disk about an axis through its center and perpendicular to the plane, we use the formula:

\[ I = \frac{1}{2} M R^{2} \]

  • Here, \( M \) represents the mass of the disk.
  • \( R \) is the radius of the disk.
This formula shows the direct relationship between the disk's geometry and its rotational properties.
Center of Mass
The center of mass of an object is the point where the entire mass of the body can be thought to be concentrated. For symmetric objects like a uniform disk, the center of mass lies exactly at its geometric center.

In the case of a uniform disk, the center of mass is crucial in determining how the body will react to external forces, such as rotation. When a disk rotates, it does so around its center of mass, contributing significantly to the calculation of its moment of inertia.

Even if a part of the disk is removed, as in this exercise, the center of mass helps to understand how the mass has been redistributed. It directly influences the new moment of inertia once a part is removed, because the symmetry and mass distribution are altered.
Axis of Rotation
The axis of rotation is an imaginary line around which an object rotates. For the initial uniform disk, this axis is perpendicular to the plane and passes through the disk's center.

This axis is crucial because it affects the distribution of mass around it, and consequently, the moment of inertia. In general, the farther the mass is from the axis of rotation, the larger the moment of inertia becomes.

In this exercise, even after a circular piece is removed from the disk, the axis of rotation remains at the center. This implies that we have to adjust the remaining distributed mass around this axis to achieve the desired change in the moment of inertia.

By understanding how the orientation and position of the axis of rotation influence moment of inertia, students are able to tackle problems involving rotational dynamics more effectively.
Circular Piece Removal
Removing a circular piece from the center of a disk might sound strange at first, but it is a classic method to alter its moment of inertia. In this exercise, a circular section is cut out to halve the disk's moment of inertia.

The moment of inertia of the removed part is important. It's calculated using the smaller radius of the piece cut out. For this, the mass of the removed portion is considered in relation to its original mass and size.

For calculation,
  • You first determine the mass of the removed section as a fraction of the whole.
  • Then, calculate its moment of inertia using the formula for disks, adjusted for its smaller size and specific radius.
This removal results in a redistribution of mass in the remaining disk, affecting the overall moment of inertia. Calculating the precise size of the piece to be removed is key to achieving the desired outcome, showcasing practical applications of physics in problem-solving.

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Most popular questions from this chapter

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?

A uniform sphere with mass \(M\) and radius \(R\) is rotating with angular speed \(\omega_{1}\) about a frictionless axle along a diameter of the sphere. The sphere has rotational kinetic energy \(K_{1}\). A thin-walled hollow sphere has the same mass and radius as the uniform sphere. It is also rotating about a fixed axis along its diameter. In terms of \(\omega_{1},\) what angular speed must the hollow sphere have if its kinetic energy is also \(K_{1},\) the same as for the uniform sphere?

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius \(12.0 \mathrm{~cm} .\) If the angular speed of the front sprocket is 0.600 rev \(/ \mathrm{s},\) what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be \(5.00 \mathrm{~m} / \mathrm{s} ?\) The rear wheel has radius \(0.330 \mathrm{~m}\).

Wheel \(A\) has three times the moment of inertia about its axis of rotation as wheel \(B .\) Wheel \(B\) 's angular speed is four times that of wheel \(A\). (a) Which wheel has the greater rotational kinetic energy? (b) If \(K_{A}\) and \(K_{B}\) are the rotational kinetic energies of the wheels, what is \(K_{A} / K_{B} ?\)

At \(t=3.00 \mathrm{~s}\) a point on the rim of a \(0.200-\mathrm{m}\) -radius wheel has a tangential speed of \(50.0 \mathrm{~m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude \(10.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00 \mathrm{~s}\) and \(t=0 .\) (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00 \mathrm{~s} ?\) (d) At what time will the radial acceleration equal \(g ?\)
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