Download the App!

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

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Determine by direct integration the moment of inertia of the shaded area with respect to the $y$ axis.

Physics 101 Mechanics

Chapter 9

Distributed Forces: Moments of Inertia

Section 1

Moments of Inertia of Areas

Moment, Impulse, and Collisions

Cornell University

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

03:30

In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

02:03

Determine by direct integr…

01:01

00:45

00:50

01:06

01:08

were asked to determine by direct integration the second moment of area of the shaded area of respect to the why access That area is bordered by a function y of X equals B times one minus k come to square root of X. We know that at exit was a Why is zero. So that tells us that K equals one over square it of a we need to integrate our function. Um, to get we need to integrate X squared all over the area of interest in that area. Why goes from minus y of X below to why of X above and then X goes from zero to a we can do the inner girls and once we crank through that, we get that I supply is to over 21 times a cube, be

View More Answers From This Book

Find Another Textbook

10:41

The T-shaped bracket shown is supported by a small wheel at $E$ and pegs at …

06:52

For the semiannular area of Prob. $5.12,$ determine the ratio $r_{1}$ to…

06:02

Four signs are mounted on a frame spanning a highway, and themagnitudes …

07:29

Knowing that $d_{C}=3 \mathrm{m},$ determine $(a)$ the distances $d_{B}$ and…

05:05

Based on experimental observations, the acceleration of a particle is define…

04:43

A lever $A B$ is hinged at $C$ and attached to a control cable at $A .$ If t…

08:40

In Prob. 10.40 , determine whether each of the positions of equilibrium is s…

08:34

Determine the magnitude and the point of application of the smallestaddi…

05:41

Determine by direct integration the centroid of the areashown. Express y…

09:52

Two uniform rods, each of mass $m$ and length $l,$ are attached to drums tha…