Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 7

Find the principal moments of inertia of a flat rectangular plate of mass \(30 \mathrm{~g}\) and dimensions \(80 \mathrm{~mm} \times 60 \mathrm{~mm}\). Given that the plate is rotating about a diagonal with angular velocity \(15 \mathrm{rad} \mathrm{s}^{-1}\), find the components of the angular momentum parallel to the edges. Given that the axis is of total length \(120 \mathrm{~mm}\), and is held vertical by bearings at its ends, find the horizontal component of the force on each bearing.

Short Answer

Expert verified
To solve this problem, we need to follow these steps: 1. Calculate the area moment of inertia about the x- and y-axis using the given mass and dimensions of the plate. 2. Find the principal moments of inertia using the formula \(I_p = (\sqrt{I_x^2 + I_y^2})\). 3. Determine the angle \(\theta\) between the diagonal axis and the x-axis, and find the components of angular velocity, \(\omega_x\) and \(\omega_y\). 4. Calculate the components of angular momentum parallel to the edges, \(L_x\) and \(L_y\), using the principal moments of inertia and the components of angular velocity. 5. Find the horizontal components of the force on each bearing, \(F_x\) and \(F_y\), using the calculated values of angular momentum and the given length of the axis. Keep in mind that these steps are a guide to the calculations, and it's essential to plug in the correct values and use the appropriate formulas to get an accurate answer.

Step by step solution

01

Calculating the Area Moment of Inertia

First, we need to find the area moment of inertia for the flat rectangular plate. For a rectangular plate, the moment of inertia about the x- and y-axis can be found using the formulas: \(I_x = (1/12)mb^2\) \(I_y = (1/12)ma^2\) Where \(m\) is the mass of the plate, \(a\) is the width, and \(b\) is the length. We are given the mass and dimensions of the plate, so we can plug those in and find the moments of inertia about the x- and y-axis.
02

Principal Moments of Inertia

Now, we can find the principal moments of inertia. The principal moments of inertia for a rectangle are given by the formula: \(I_p = (\sqrt{I_x^2 + I_y^2})\) We can plug in the values of \(I_x\) and \(I_y\) we found in Step 1 to find the principal moments of inertia, \(I_p\).
03

Angular Velocity Components

We are given the angular velocity of the plate, which is 15 rad/s. Since the plate is rotating about a diagonal, the components of angular velocity parallel to the edges can be found as follows: \(\omega_x = \omega \cos(\theta)\) \(\omega_y = \omega \sin(\theta)\) Where \(\theta\) is the angle between the diagonal axis and the x-axis. For a rectangle, this angle can be found using the tangent function: \(\tan(\theta) = \frac{b}{a}\) Calculate the angle \(\theta\), and then plug in the given values of \(\omega\) to find the components of angular velocity, \(\omega_x\) and \(\omega_y\).
04

Angular Momentum Components

Now, we can find the components of angular momentum parallel to the edges of the plate using the following formulas: \(L_x = I_p\omega_x\) \(L_y = I_p\omega_y\) Plug in the values of \(I_p\), \(\omega_x\), and \(\omega_y\) we found in the previous steps, to find the components of angular momentum, \(L_x\) and \(L_y\).
05

Horizontal Components of Force

Given that the axis is of total length 120 mm, and the force is distributed across two bearings on either end, we can find the horizontal component of the force on each bearing using the following formulas: \(F_x = \frac{L_x^2 + L_y^2}{2l}\) \(F_y = -\frac{L_x^2 + L_y^2}{2l}\) Where \(l\) is the length of the axis (120 mm). Plug in the values of \(L_x, L_y\), and \(l\) to find the components of the force on each bearing, \(F_x\) and \(F_y\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 solid rectangular box, of dimensions \(100 \mathrm{~mm} \times 60 \mathrm{~mm} \times 20 \mathrm{~mm}\), is spinning freely with angular velocity 240 r.p.m. Determine the frequency of small oscillations of the axis, if the axis of rotation is (a) the longest, and (b) the shortest, axis.

A rigid body of spheroidal shape, spinning rapidly about its axis of symmetry, is placed on a smooth flat table. Show by considering the moment of the force at the point of contact that its axis will precess in one direction if it is oblate \((ca=b, e . g\). a rugby ball). Show also that if there is a small frictional force, the axis will become more nearly vertical, so that if the body is oblate its centre of mass will fall, but if it is prolate it will rise.

Find the moment of inertia about an axis through its centre of a uniform hollow sphere of mass \(M\) and outer and inner radii \(a\) and \(b\). (Hint: Think of it as a sphere of density \(\rho\) and radius \(a\), with a sphere of density \(\rho\) and radius \(b\) removed.)

A uniform rod of mass \(M\) and length \(2 a\) hangs from a smooth hinge at one end. Find the length of the equivalent simple pendulum. It is struck sharply, with impulse \(X\), at a point a distance \(b\) below the hinge. Use the angular momentum equation to find the initial value of the angular velocity. Find also the initial momentum. Determine the point at which the rod may be struck without producing any impulsive reaction at the hinge. Show that, if the rod is struck elsewhere, the direction of the impulsive reaction depends on whether the point of impact is above or below this point.

Calculate the principal moments of inertia of a uniform, solid cone of vertical height \(h\), and base radius \(a\) about its vertex. For what value of the ratio \(h / a\) is every axis through the vertex a principal axis? For this case, find the position of the centre of mass and the principal moments of inertia about it.
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Geometrical and Physical Optics

Read Explanation

Linear Momentum

Read Explanation

Medical Physics

Read Explanation

Translational Dynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation
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.

Sign-up for free