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A thin wire of constant linear mass density $k$ takes the shape of an arch of the cycloid $$x = a(t − \sin t),\quad y = a(1 − \cos t), \quad 0 ≤ t ≤ 2π.$$

Determine the mass $m$ of the wire, and find the location of its center of mass.

I am assuming I am supposed to use line integrals and vector fields to solve this. Any help would be appreciated. :)

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3 Answers 3

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Hints:

To find the mass, find the arc-length and multiply with the density.

To find the center of mass, you know that, due to symmetry, it must lie halfway through the arc in the $x$-direction. Now you only need to find the mean value of the $y$-values present in the arc.

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  • $\begingroup$ To find the arc length, you find x' and y', square, add, put under a square root, and then integrate from 0 to 2pi right? $\endgroup$
    – Jack Katt
    Commented May 5, 2017 at 3:00
  • $\begingroup$ @idk Yes, exactly! :) $\endgroup$ Commented May 5, 2017 at 8:33
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The object is to find the arc length $s$ of the curve

$$x = a(t − \sin t),\quad y = a(1 − \cos t), \quad 0 ≤ t ≤ 2π$$

We can set $a=1$ since it's just a scaling factor.

The arc length is given by

$$s=\int_0^{2\pi}\sqrt{1+\left(\frac{dy}{dx} \right)^2 }\ dx$$

Now, we have

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\sin t}{1-\cos t}$$

and

$$s=\int_0^{2\pi}\sqrt{1+\left(\frac{dy}{dx} \right)^2 }\ \frac{dx}{dt}dt$$

After some straightforward manipulation we obtain

$$s=\int_0^{2\pi}\sqrt{(1-\cos t)^2+\sin^2 t}\ dt=\int_0^{2\pi}\sqrt{2-2\cos t}\ dt$$

With the help of WolframAlpha we get

$$s=2\sqrt{2-2\cos t}\cdot\cot(t/2)\big|_0^{2\pi}=8$$

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You have a 1D domain, the position vector $\underline{{\rm pos}}(t) = \pmatrix{x(t)&y(t)&z(t)}$ in terms of a single parameter $t$.

To integrate over this curve, you need a description of how much length ${\rm d}s$ is traversed for each change in parameter ${\rm d}t$

Using differentiation you find the velocity vector at each location as

$$\underline{{\rm vel}}(t) = \frac{\partial}{\partial t} \underline{{\rm pos}}(t) $$

The speed at each location is then found using the vector magnitude of the length

$$ {\rm speed}(t) = \sqrt{ \underline{{\rm vel}}(t) \cdot \underline{{\rm vel}}(t) } = \sqrt{ \dot{x}^2 + \dot{y}^2 + \dot{z}^2}$$

where $\cdot$ is the vector dot product, and $\dot{x} = \frac{\partial}{\partial t}x$ is rate of change in $x$, and similarly for the other coordinates.

The above in integral form gives us the length differential

$${\rm d}s = \sqrt{ \underline{{\rm vel}}(t) \cdot \underline{{\rm vel}}(t) } \,{\rm d}t = \sqrt{ \dot{x}^2 + \dot{y}^2 + \dot{z}^2}\; {\rm d}t$$

Not to perform the integral, use the expressions below

  • Length $$ \ell = \int \sqrt{ \dot{x}^2 + \dot{y}^2 + \dot{z}^2}\; {\rm d}t$$

  • Mass $$ m = \int k \sqrt{ \dot{x}^2 + \dot{y}^2 + \dot{z}^2}\; {\rm d}t $$

  • Center of Mass $$\underline{\rm com} = \frac{1}{m} \int \underline{\rm pos}(t) k \sqrt{ \dot{x}^2 + \dot{y}^2 + \dot{z}^2}\; {\rm d}t $$

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