Question

Find the mass of a thin wire which is in the shape of a helix and center of mass of a thin wire which is in the shape of a helix.

Short answer

The mass of a helix shaped thin wire is \(\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)\) and center of mass of a helix shaped thin wire is \(\left( {\frac{{3\pi \left( {2{\pi ^2} + 1} \right)}}{{\left( {4{\pi ^2} + 3} \right)}},\frac{6}{{\left( {4{\pi ^2} + 3} \right)}}, - \frac{{12\pi }}{{\left( {4{\pi ^2} + 3} \right)}}} \right)\).

Step-by-step solution

Given data

The parametric equations of the helix shaped thin wire are\(x = t,y = \cos t,z = \sin t,0 \le t \le 2\pi \). The density function of the thin wire is the square of the distance from the origin.

Concept used

The mass of a thin wire which is in the shape of a curve\(C\)by the use of the equation as follows:\(m = \int_C \rho (x,y)ds\).

Here,\(\rho (x,y)\)is the linear density and\(ds\)is the change in surface.

Write the formulae for the center of mass of a thin wire which is in the shape of a space curve

\(C\)

Write the expression for the mass of the thin wire as follows:

\(m = \int_C \rho (x,y,z)ds\) ……. (1)

Write the expression for \(ds\) as follows:

\(ds = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} dt\) ……. (2)

Write the expression to find density function as follows:

\(\rho (x,y,z) = {d^2}\) ……. (3)

Here, \(d\) is the distance between any point to the wire to the origin.

Write the expression to find distance between two points \(\left( {{x_1},{y_1},{z_1}} \right)\) and \(\left( {{x_2},{y_2},{z_2}} \right)\) as follows:

\(d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \) ……. (4)

Write the expression to find the \(x\)-coordinate of a center of mass of the thin wire as follows: \(\bar x = \frac{1}{m}\int_C x \rho (x,y,z)ds\). ……. (5)

Formula for the \(y\)-coordinate of center of mass of the thin wire as follows:

\(\bar y = \frac{1}{m}\int_C y \rho (x,y,z)ds\) ……. (6)

Formula for the \(z\)-coordinate of center of mass of the thin wire as follows:

\(\bar z = \frac{1}{m}\int_C z \rho (x,y,z)ds\) ……. (7)

Consider a point \((x,y,z)\) on the thin wire.

As the density function \(\rho (x,y,z)\) is the square of the distance between any point on the wire to the origin, find the distance between the point \((x,y,z)\) to the origin as follows:

Substitute \(x\) for \({x_2},y\) for \({y_2},z\) for \({z_2},0\) for \({x_1},0\) for \({y_1}\), and 0 for \({z_1}\) in equation (4) as follows:

\(\begin{array}{c}d = \sqrt {{{(x - 0)}^2} + {{(y - 0)}^2} + {{(z - 0)}^2}} \\ = \sqrt {{x^2} + {y^2} + {z^2}} \end{array}\)

Calculation of density function \(\rho (x,y,z)\) as follows:

Substitute \(\sqrt {{x^2} + {y^2} + {z^2}} \) for \(d\) in equation (3) as follows:

\(\begin{array}{c}\rho (x,y,z) = {\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)^2}\\ = {x^2} + {y^2} + {z^2}\end{array}\)

Substitute \(t\) for \(x,\cos ty\), and \(\sin t\) for \(z\) as follows:

Calculation of \(ds\)

Substitute \(t\) for \(x,\cos t\) for \(y\), and \(\sin t\) for \(z\) in equation (2) as follows:

\(\begin{aligned}ds & = \sqrt {{{\left( {\frac{d}{{dt}}t} \right)}^2} + {{\left( {\frac{d}{{dt}}\cos t} \right)}^2} + {{\left( {\frac{d}{{dt}}\sin t} \right)}^2}} dt\\ & = \sqrt {{{(1)}^2} + {{( - \sin t)}^2} + {{(\cos t)}^2}} dt\\ & = \sqrt {1 + {{\sin }^2}t + {{\cos }^2}t} dt\quad \end{aligned}\)

Simplify the expression as follows:

\(\begin{aligned}ds &= \sqrt {1 + (1)} dtds\\ &= \sqrt 2 dt\end{aligned}\)

Calculation of \(m\)

Substitute \(\left( {{t^2} + 1} \right)\) for \(\rho (x,y,z),\sqrt 2 dt\) for \(ds,0\) for lower limit, and \(2\pi \) for upper limit in equation (1) as follows:

\(\begin{aligned}m &= \int_0^{2\pi } {\left( {{t^2} + 1} \right)} (\sqrt 2 dt)\\ &= \sqrt 2 \int_0^{2\pi } {\left( {{t^2} + 1} \right)} dt\\ & = \sqrt 2 \left[ {\frac{{{t^3}}}{3} + t} \right]_0^{2\pi }\\ & = \sqrt 2 \left\{ {\left[ {\frac{{{{(2\pi )}^3}}}{3} + 2\pi } \right] - \left( {\frac{{{0^3}}}{3} + 0} \right)} \right\}\end{aligned}\)

Simplify the expression as follows:

\(\begin{aligned}m & = \sqrt 2 \left[ {\left( {\frac{{8{\pi ^3}}}{3} + 2\pi } \right) - 0} \right]\\ & = \sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)\end{aligned}\)

Thus, the mass of a helix shaped thin wire is \(\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)\).

Calculation of \(\bar x\)

Substitute \(\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)\) for \(m,t\) for \(x,\left( {{t^2} + 1} \right)\) for \(\rho (x,y,z),\sqrt 2 dt\) for \(ds,0\) for lower limit, and \(2\pi \) for upper limit in equation (5) as follows:

\(\begin{aligned}\bar x & = \frac{1}{{\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)}}\int_0^{2\pi } {(t)} \left( {{t^2} + 1} \right)(\sqrt 2 dt)\\ & = \frac{{\sqrt 2 }}{{\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)}}\int_0^{2\pi } {\left( {{t^3} + t} \right)} dt\\ & = \frac{1}{{\left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)}}\left[ {\frac{{{t^4}}}{4} + \frac{{{t^2}}}{2}} \right]_0^{2\pi }\\ & = \frac{3}{{4\left( {8{\pi ^3} + 6\pi } \right)}}\left[ {{t^4} + 2{t^2}} \right]_0^{2\pi }\end{aligned}\)

Simplify the expression as follows:

\(\begin{aligned}\bar x & = \frac{3}{{4\left( {8{\pi ^3} + 6\pi } \right)}}\left\{ {\left[ {{{(2\pi )}^4} + 2{{(2\pi )}^2}} \right] - \left[ {{0^4} + 2{{(0)}^2}} \right]} \right\}\\ & = \frac{3}{{4\left( {8{\pi ^3} + 6\pi } \right)}}\left( {16{\pi ^4} + 8{\pi ^2}} \right)\\ & = \frac{{3\left( {8{\pi ^2}} \right)\left( {2{\pi ^2} + 1} \right)}}{{4(2\pi )\left( {4{\pi ^2} + 3} \right)}}\\ & = \frac{{3\pi \left( {2{\pi ^2} + 1} \right)}}{{\left( {4{\pi ^2} + 3} \right)}}\end{aligned}\)

Calculation of \(\bar y\)

Substitute \(\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)\) for \(m,\cos t\) for \(y,\left( {{t^2} + 1} \right)\) for \(\rho (x,y,z),\sqrt 2 dt\) for \(ds,0\) for lower limit, and \(2\pi \) for upper limit in equation (6) as follows:

\(\begin{aligned}\bar y & = \frac{1}{{\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)}}\int_0^{2\pi } {(\cos t)} ({t^2} + 1)(\sqrt 2 dt)\\ & = \frac{{\sqrt 2 }}{{\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)}}\int_0^{2\pi } {({t^2}\cos t + \cos t)} dt\\ & = \frac{3}{{\left( {8{\pi ^3} + 6\pi } \right)}}\left[ {\int\limits_0^{2\pi } {\left( {{t^2}\cos t} \right)dt + \int\limits_0^{2\pi } {\left( {\cos t} \right)dt} } } \right]\\ & = \frac{3}{{\left( {8{\pi ^3} + 6\pi } \right)}}\left[ {\left( {{t^2} - 2} \right)\sin t + 2t\cos t} \right]_0^{2\pi } + \left[ {\sin t} \right]_0^{2\pi }\end{aligned}\)

Simplify the expression as follows:

\(\begin{aligned}\bar y & = \frac{3}{{\left( {8{\pi ^3} + 6\pi } \right)}}\left( {\begin{aligned}{*{20}{l}}{\left\{ {\begin{aligned}{*{20}{l}}{\left. {\left[ {{{(2\pi )}^2} - 2} \right)\sin (2\pi ) + 2(2\pi )\cos (2\pi )} \right]}\\{ - \left[ {\left( {{0^2} - 2} \right)\sin (0) + 2(0)\cos (0)} \right]}\\{ + [\sin (2\pi ) - \sin (0)]}\end{aligned}} \right\}}\end{aligned}} \right)\\ & = \frac{3}{{\left( {8{\pi ^3} + 6\pi } \right)}}\left( {\left\{ {\left[ {\left( {4{\pi ^2} - 2} \right)(0) + 4\pi (1)} \right] - 0} \right\} + 0} \right)\\ & = \frac{{3(4\pi )}}{{(2\pi )\left( {4{\pi ^2} + 3} \right)}}\\ & = \frac{6}{{\left( {4{\pi ^2} + 3} \right)}}\end{aligned}\)

Calculation of \(\bar z\)

Substitute \(\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)\) for \(m\), \(\sin t\) for \(z,\left( {{t^2} + 1} \right)\) for \(\rho (x,y,z),\sqrt 2 dt\) for \(ds,0\) for lower limit, and \(2\pi \) for upper limit in equation ( 7 ) as follows:

\(\begin{aligned}\bar z & = \frac{1}{{\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)}}\int_0^{2\pi } {(\sin t)} ({t^2} + 1)(\sqrt 2 dt)\\ & = \frac{{\sqrt 2 }}{{\sqrt 2 \left( {\frac{8}{3}{\pi ^3} + 2\pi } \right)}}\int_0^{2\pi } {({t^2}\sin t} + \sin t)dt\\ & = \frac{3}{{8{\pi ^3} + 6\pi }}\left[ {\int_0^{2\pi } {({t^2}\sin t} )dt + \int\limits_0^{2\pi } {(\sin t)} dt} \right]\\ & = \frac{3}{{8{\pi ^3} + 6\pi }}\left[ {(2t\sin t - ({t^2} - 2)\cos t} \right]_0^{2\pi } - \left[ {\cos t} \right]_0^{2\pi }\end{aligned}\)

Simplify the expression as follows:

\(\begin{aligned}\bar z & = \frac{3}{{\left( {8{\pi ^3} + 6\pi } \right)}} - \left[ {2(2\pi )\sin \;(2\pi ) - \left( {{{(2\pi )}^2} - 2} \right)\cos \;(2\pi ) - [2(0)\sin \;(0) - \left( {{{(0)}^2} - 2} \right)\cos \;(2(0))] - [\cos \;(2\pi ) - \cos \;(0)]} \right]\\ & = \frac{3}{{\left( {8{\pi ^3} + 6\pi } \right)}}\left( {\left\{ {\left[ {4\pi (0) - \left( {4{\pi ^2} - 2} \right)(1)} \right] - [ - ( - 2)(1)]} \right\} - [1 - 1]} \right)\\ & = \frac{3}{{(2\pi )\left( {4{\pi ^2} + 3} \right)}}\left( { - 4{\pi ^2} + 2 - 2} \right)\\ & = - \frac{{6\left( {4{\pi ^2}} \right)}}{{(2\pi )\left( {4{\pi ^2} + 3} \right)}}\end{aligned}\)

Thus, the center of mass of a helix shaped thin wire is: \(\left( {\frac{{3\pi \left( {2{\pi ^2} + 1} \right)}}{{\left( {4{\pi ^2} + 3} \right)}},\frac{6}{{\left( {4{\pi ^2} + 3} \right)}}, - \frac{{12\pi }}{{\left( {4{\pi ^2} + 3} \right)}}} \right)\).