# SOLVED: Generalization of the parallel-axis theorem to the inertia tensor Let I(CM) denote the moment of inertia tensor for an arbitrary rigid body rotating about its center of mass (CM). Let I(CM) de

Understanding the Area Moment of Inertia
Understanding the Area Moment of Inertia

Get 5 free video unlocks on our app with code GOMOBILE

Snapsolve any problem by taking a picture.
Try it in the Numerade app?

Generalization of the parallel-axis theorem to the inertia tensor
Let I(CM) denote the moment of inertia tensor for an arbitrary rigid body rotating about its center of mass (CM). Let I(CM) denote the inertia tensor for rotations about a point displaced from the CM by an amount A-(KZ). Prove the following results:
I(CM) = I(CM) + M(Y? IYCM) + M(X? 2-) I(CM) + M(X? Y2) I(CM) + MXY I(CM) + MXZ I(CM) + MYZ Iyy
Show all your work in full detail.

This problem has been solved!

Try Numerade free for 7 days

01:36

Prove the “parallel axis theorem “: The moment of inertia $I$ of a body about a given axis is $I=I_{m}+M d^{2}$, where $M$ is the mass of the body, $I_{m}$ is the moment of inertia of the body about an axis through the center of mass and parallel to the given axis, and $d$ is the distance between the two axes.

01:38

Parallel axis theorem. Beginning with the fact that the moment of inertia of a thin disk about a diametral axis is $\frac{1}{4} m a^{2}$, employ the parallel axis theorem to prove that for a solid circular cylinder of mass $M$, radius $a$, and length $L$, the moment of inertia about a transverse axis through the center of mass is $M a^{2} / 4+M L^{2} / 12$

01:00

The parallel axis theorem provides a useful way to calculate the moment of inertia $I$ about an arbitrary axis. The theorem states that $I=I_{\mathrm{cm}}+M h^{2},$ where $I_{\mathrm{cm}}$ is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, M is the total mass of the object, and h is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius R relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

06:40

[Parallel-Axis Theorem] In Lecture WC, we directly calculated the moment of inertia of a dumbbell consisting of two small masses (m) at the ends of a massless stick of length (D). We did this for both an axis perpendicular to the stick through the center of mass (CM) and for an axis through one end. This will show how to use the Parallel-Axis Theorem to find the moment of inertia (Iend) about an axis through one end, using the moment of inertia (ICM) about an axis through the CM. Compare your result with the result obtained by direct calculation of (Iend).

Oops! There was an issue generating an instant solution

An exclusive promotional rate that’s yours to keep forever

or

EMAIL