Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

(III) (a) Derive the formula given in Fig. 20 $\mathrm{h}$ for the moment of inertia of a uniform, flat, rectangular plate of dimensions $\ell \times w$ about an axis through its center, perpendicular to the plate. (b) What is the moment of inertia about each of the axes through the center that are parallel to the edges of the plate?

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

(A) $\mathrm{I}=\frac{1}{12} \mathrm{M}\left(1^{2}+\mathrm{w}^{2}\right)$(b) $\mathrm{I}_{\mathrm{x}}=\frac{1}{12} \mathrm{M} \mathrm{w}^{2}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Washington

Hope College

University of Sheffield

Lectures

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

10:07

Calculate the moment of in…

06:05

00:35

The moment of inertia for …

02:28

(III) Derive the formula f…

04:05

(a) For the thin rectangul…

01:46

Let $I$ be the moment of i…

06:52

So here we can say that we're going to choose a differential area equaling d x times d y. And this would essentially divide the plate up into differential rectangular elements where each element has a differential area. So we can say that the mass of an element ah d m would be equaling. I am over the length times, though with multiplied bye d x d y ah, the distance of that element from the axis of rotation are, we can say is equaling the square root of X squared plus y squared. And we're going to then use Equation 10 16 to calculate the moment of inertia. So we can say that the moment of inertia of the center would be equaling square root of R squared times D. Mm. This would essentially be equaling withy integral from negative with over two to the with over two times the integral from the negative out the learned native length of A to tow the length over too, multiplied by X squared plus y squared multiplied by em over the length times the with times d x d y. And so we can then say that the moment of inertia through the center. Ah would be equaling four em over the length times the with times, uh, zero to the with over two times the integral from zero to the length over, too. Times X squared plus y squared times, dx, dy y This would Essentially, we can solve one part on. Say that here the moment of inertia. Now that we can just say this would be equal to four em over al W times the integral from zero to the with over two multiplied by 1/3 times 1/2 of the length we can say 1/3 time's the length over to quantity cubed plus 1/2. Rather, we can say the length over too times why squared times D y. And so the moment of inertia through the center would be equaling two times over W times the integral from zero to the with over two multiplied by l squared over 12. Plus why squared times D y and this is gonna be equaling to, um over w times l squared over 12 multiplied by the with over two plus 1/3 times, one the with over two rather quantity cube and we finally find that the moment of inertia through the center was equaling and over 12 multiplied by the lake, the length squared plus the with squared. And so this would be our final answer. Four part eh and then four part B we ascend. We have here for the axis of rotation parallel to the with dimension. Essentially the acts the rotation access would be in the UAE direction. We can consider the plate to be made out of a large number of thin rods. Uh, each rod has its length l and it's rotating about an access through its center. So we can say that the moment of inertia of one rod would be equaling 1/12. Uh um sub I l squared, and absolutely I would be the mass of one single rot. So for a collection of these rods, we can say that the moment of inertia with the rotation access being in the UAE direction, this would be equaling the some starting II of 1 12 1 12 um, initial I'll square and this is simply equaling 1/12 times the total mass at times l squared. Similarly, if the axis of rotation was for in rather in the X axis ice of X. This would be equaling 1/12 times the total mass of the collection of rods. But now, instead of the length squared, this would be the with squared, and this essentially proves the pair of the perpendicular access serum. So with this combined with this, this is essentially proving equation 10 18 where the moment of inertia through the Z axis would be equaling the moment of inertia through the X plus the moment of inertia through the why. And that's essentially what we calculated here that would be our answer for Part B. That is the end of the solution. Thank you for watching.

View More Answers From This Book

Find Another Textbook