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Chapter 9: Problem 23

If the moment of inertia of a dise about an axis tangential and parallel to its surface be \(\mathrm{I}\), then what will be the moment of inertia about the axis tangential but perpendicular to the surface? (a) \(\frac{6}{5} I\) (b) \(\frac{3}{4}\) ? (c) \(\frac{3}{2}\), (d) \(\left.\frac{5}{4}\right\\}\)

Short Answer

Expert verified
(c) \(\frac{3}{2}\)

Step by step solution

01

Understanding the Parallel Axis Theorem

The parallel axis theorem states that the moment of inertia, \(I_{ ext{new}}\), about any axis parallel to an axis through the center of mass is given by the formula: \(I_{ ext{new}} = I_{ ext{cm}} + md^2\), where \(I_{ ext{cm}}\) is the moment of inertia about the center of mass axis, \(m\) is the mass of the object, and \(d\) is the perpendicular distance between the axes.
02

Applying Parallel Axis Theorem for Given Axis

According to the problem, the given moment of inertia \(I\) is about an axis that is tangential and parallel to the surface. This means \(I = I_{ ext{cm}} + mR^2\), where \(R\) is the radius of the disc.
03

Find \(I_{cm}\) for the Disc

The moment of inertia of a disc about an axis through its center and perpendicular is \(I_{cm} = \frac{1}{2}mR^2\). This is a standard result for discs.
04

Calculate the Moment of Inertia for Perpendicular Axis

We wish to find the moment of inertia about an axis perpendicular to the surface, which is also tangential. Using the parallel axis theorem, we have: \(I_{ ext{perpendicular}} = I_{cm} + md^2\). Here, \(d = R\), so \(I_{ ext{perpendicular}} = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2\).
05

Relate to Initial Given Inertia

From the problem statement, we know \(I = I_{cm} + mR^2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2\). Therefore, the moment of inertia about the axis perpendicular and tangential is the same as \(I\), meaning: \(I_{\text{perpendicular}} = I\).

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Key Concepts

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

Parallel Axis Theorem
The Parallel Axis Theorem is a crucial concept in understanding how the moment of inertia changes when an object is rotated about an axis that is not its center of mass. The theorem states: \( I_{\text{new}} = I_{\text{cm}} + md^2 \). Here, \( I_{\text{new}} \) represents the moment of inertia around the new axis, \( I_{\text{cm}} \) is the moment of inertia around an axis through the center of mass, \( m \) is the mass of the object, and \( d \) is the perpendicular distance between the center of mass axis and the new axis.

  • This theorem allows us to calculate the moment of inertia around any axis parallel to a known axis of an object.
  • It is particularly useful when dealing with composite bodies or bodies with complicated mass distributions.
  • The theorem simplifies the calculation by considering the distribution of mass relative to the axis of rotation.
Understanding this theorem is essential for solving problems involving rotating discs, rods, and other shapes where axes change.
Disc
A disc is a flat, circular object that has rotational symmetry around its central axis. The moment of inertia for a disc depends on its axis of rotation. Typically, the moment of inertia about the axis through its center and perpendicular to its surface is particularly important in calculations.
  • The standard moment of inertia for a disc having mass \( m \) and radius \( R \) about its central perpendicular axis is: \( I_{\text{cm}} = \frac{1}{2}mR^2 \).
  • This property makes discs handy to model in mechanical systems, like wheels or flywheels, due to symmetrical mass distribution.
  • When calculating moment of inertia about different axes, the symmetry often simplifies the use of the Parallel Axis Theorem.
For understanding the inertia of a disc about different axes, it's crucial to start with its primary inertia about the central axis.
Tangential Axis
A tangential axis is an axis that lies in the plane of an object and touches it at only one point, essentially being tangent to the surface of the object. Calculating the moment of inertia around this axis can be more complex as it does not pass through the object's center of mass.
  • To determine inertia around a tangential axis, we often use the Parallel Axis Theorem.
  • In a disc, moving from a central to a tangential axis involves increasing the moment of inertia because the mass distribution's leverage around the axis increases.
  • If the axis is also parallel to a known axis, such as the central axis, calculations are straightforward using the theorem.
For a disc, this shift in axis significantly affects the resulting moment of inertia calculations, as shown in physics problems.
Perpendicular Axis
The Perpendicular Axis Theorem is another vital concept, though it applies specifically to planar objects like discs. For a planar body lying in the \( x-y \) plane, if \( I_x \) and \( I_y \) are the moments of inertia about two perpendicular axes in the plane, then the moment of inertia about an axis perpendicular to this plane, passing through their intersection (center), is given by the sum: \[ I_z = I_x + I_y \]This relates directly to how moments of inertia are distributed across different orientations in space.
  • This theorem simplifies the calculation when needing the 3D moment of inertia from two-dimensional axes.
  • For symmetrical shapes like discs, this can provide insights into their rotational behavior in different planes.
  • Though it doesn鈥檛 directly apply to the exercise, understanding this theorem enhances the comprehension of rotational dynamics for 2D bodies.
Applying the Perpendicular Axis Theorem alongside other principles extends the understanding of rotational inertia in complex scenarios.

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Most popular questions from this chapter

Consider a uniform square plate of side \(a\) and mass \(m\). The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is (a) \(\frac{5}{6} m a^{2}\) (b) \(\frac{m a^{2}}{12}\) (c) \(\frac{7}{12} m a^{2}\) (d) \(\frac{2}{3} m a^{2}\)

The oxygen molecule has a mass of \(5.30 \times 10^{-26} \mathrm{~kg}\) and a moment of inertia of \(1.94 \times 10^{-46} \mathrm{~kg}-\mathrm{m}^{-2}\) about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is \(500 \mathrm{~m} / \mathrm{s}\) and that is KE of rotation is \(\frac{2}{3}\) of its KE translation. Find the average angular velocity of the molecule. (a) \(3.75 \times 10^{12} \mathrm{rad} / \mathrm{s}\) (b) \(5.75 \times 10^{12} \mathrm{rad} / \mathrm{s}\) (c) \(9.75 \times 10^{12} \mathrm{rad} / \mathrm{s}\) (d) \(6.75 \times 10^{12} \mathrm{rad} / \mathrm{s}\)

The moment of inertia of a dumb-bell, consisting of point masses \(m_{1}=2.0 \mathrm{~kg}\) and \(m_{2}=1.0 \mathrm{~kg}\), fixed to the ends of a rigid massless rod of length \(L=0.6 \mathrm{~m}\), about an axis passing through the centre of mass and perpendicular to its length, is (a) \(0.72 \mathrm{~kg} \mathrm{~m}^{2}\) (b) \(0.36 \mathrm{~kg} \mathrm{~m}^{2}\) (c) \(0.27 \mathrm{~kg} \mathrm{~m}^{2}\) (d) \(0.24 \mathrm{~kg} \mathrm{~m}^{2}\)

A thin uniform \(\operatorname{rod} A B\) of mass \(m\) and length \(L\) is hinged at one end \(A\) to the level floor. Initially, it stands vertically and is allowed to fall freely to the floor in the vertical plane. The angular velocity of the rod, when its end \(B\) strikes the floor is \((g\) is acceleration due to gravity) (a) \(\left(\frac{m g}{L}\right)\) (b) \(\left(\frac{m g}{3 L}\right)^{1 / 2}\) (c) \(\left(\frac{g}{L}\right)\) (d) \(\left(\frac{3 g}{L}\right)^{1 / 2}\)

What is the torque of the force \(\mathbf{F}=(2 \mathbf{i}-3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}) \mathrm{N}\) acting at the point \(\mathbf{r}=(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \mathrm{m}\) about the origin? (a) \(-17 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}\) (b) \(-6 \hat{i}+6 \hat{j}-12 \hat{k}\) (c) \(17 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-13 \hat{\mathbf{k}}\) (d) \(6 \hat{i}-6 \hat{j}+12 \hat{k}\)
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