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The density (mass per unit length) of a thin rod of length $\ell$ increases uniformly from $\lambda_{0}$ at one end to 3$\lambda_{0}$ at the other end. Determine the moment of inertia about an axis perpendicular to the rod through its geometric center.

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$\frac{1}{6} \lambda_{0} \ell^{3}$

Physics 101 Mechanics

Chapter 10

Rotational Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

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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.

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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.

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How could we be moment of inertia about an access through the geometric center of the Rod? So we can say that we're going to select a differential element of the rod of length DX So this would be the differential length Essentially, um, and it would be at a distance x from the center of the rod. Here. This dash line represents the essentially the axis of rotation because the mass density changes uniformly from the, uh, initial mass density, we consider it here. We can say that here acts, we have the initial mass density at Rather, we can say X equals rather negative Al over to. And then we can say that this will be three times the initial mass density at X equaling l over to. So here we can see that the mast on mass density function lambda this would be equal to two times landed multiplied by one over one mother one plus X over l. And we're essentially finding the mass of the differential elements. We can say the M is equaling landed D X. This is gonna be equal to two Lambda one plus X over out multiplied by D X and So now we're going to use the equation 10 16 to calculate the moment of inertia. So the moment of inertia about an end will be equaling the square root of our square times D M. Um, And then we can Let's actually substitute moment of inertia about an end. This would be equaling square root of negative L over to to Oliver, too. And this would be X squared times two Lambda one plus X over al the X and this is gonna equal to times lambda multiplied by the square root of negative L over two times land over to times X squared plus X cubed over l Times D X. And so we can then say that the moment of inertia I will be equaling two times Lambda multiplied by X cubed over three plus x to the fourth divided by four times l evaluated atnegative l over two and positive l over too. And so this is going to be equal after substance subbing in We have 1/6 Lambda Times l cubed. This would be our final answer for the moment of inertia. That is the end of the solution. Thank you for watching

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