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# Two springs of constant density A spring of constant density $\delta$ lies along the helix$$\mathbf { r } ( t ) = ( \cos t ) \mathbf { i } + ( \sin t ) \mathbf { j } + t \mathbf { k } , \quad 0 \leq t \leq 2 \pi$$a. Find $I _ { z }$b. Suppose that you have another spring of constant density $\delta$ that is twice as long as the spring in part (a) and lies along the helix for $0 \leq t \leq 4 \pi .$ Do you expect $I _ { z }$ for the longer spring to be the same as that for the shorter one, or should it be different? Check your prediction by calculating $I _ { z }$ for the longer spring.

## $2 \pi \sqrt{2} \delta$, $4 \pi \sqrt{2} \delta$

Integrals

Vectors

Vector Functions

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##### Lily A.

Johns Hopkins University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

we're gonna go ahead and talk about, um, a spring. So we have a spring with, um, a constant density. Um, and it lies, um, along a helix. Um, with, um, r of t is equal to co sign t I plus sign of TJ plus t k and T goes from 0 to 2 pi inclusive. And the first thing we wanted Thio is we wanna find the moment of inertia about the Z axis. Okay, so we know that is defined to be the inner Groll over the curve of X squared plus y squared, Um, times that density, Uh, with respect to d s. Okay, So what we need to do is to first of all, find out what D s is. Um, and, um, d s is equal to the magnitude of this v a t function. Um, time C T. So V of tea is actually equal to the derivative of the or function or the parametric function. So we take the derivative of that, and that gives me native sign of tea times. I place co sign of tea times J plus K. So the magnitude of that is given by the square root uh, negative sign of tea, and we're going to square that term plus co sign of tea, and we're going to square that plus one squared. Um, And so it's each of the terms of the values before I, J and K, and so a sign square because the co sign squared gives me a one. Um, and so we have the square root of two, and so d s is equal to the square to two times DT. Okay, so now what we need to do is we have that moment of inertia about the sea is equal to the integral from 0 to 2 pi of, um x squared and x was co side of tea. So that's co sign squared of tea plus y squared. So that's gonna be a sine squared of tea. Um, lips, times, um, the density times D s, which is the square root of two times t t. Okay, um and so, um, co sign squared because the sine squared gives me a one. So I have the square to two times identity times that, you know, girl from 0 to 2 pi uh, t t. And so this is going to give me the square root of two times the density Times t evaluated at two pi and zero. So this is to pie time discredited, too. Times the density. So there is a moment of inertia. Um, for that, um, spring with a constant density. And so now what we want to do is suppose, um So suppose you have another spring, and it has a constant density as well. Um, but it is twice as long, um, as this previous spring, and it lies on a helix as well. Um, on the same, he licks Not the same, but a similar helix that has that same parametric equation. But this time, um, t is going to go from 0 to 4 pi, and we want to know. Okay, Um, do you sake? The moment of inertia about the Z is the same or different. Um, and I'm gonna go ahead and say is going to be different. Um, it has a constant density. The same. So that's not gonna change. Um, twice as long as the length, Um, doesn't really matter, because we don't have the length in here, but we do know that it goes from zero instead of going from 0 to 2. Pi is gonna go from 0 to 4 pi. So I would almost guess that it's four pi Time to square 22 times that density because nothing else is changing other Thandie. The only thing that is changing or that's gonna be different appear or those, um, upper or lower limits. So that moment of inertia gonna go from 0 to 2 pi The function is not gonna change. We still have co sign squared of tea plus sine squared of tea. Um, we still have that constant density, and we still have the square to two DT because it's all based. Both of these were based on that parametric equation. And so now when we do this, um, we get once again this will be a one. And so we're bet we're basically right here. So two times that density from 0 to 4 pi now hopes this should be a four pi Um uh d t. So, um oh, times the square ready to not to, um and so this is gonna be the square root of two times that density times t from 0 to 4 pi. So now this is four pi times the square to two times that constant density. So it is thio, um, different. It's twice the moment of inertia for this one is twice the moment your show, the other one.

University of Central Arkansas

#### Topics

Integrals

Vectors

Vector Functions

##### Lily A.

Johns Hopkins University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp