Winter 2016 - Q.8 - Grad Plus

# Winter 2016 – Q.8

8. a) A particle moves so that its position rector is given by $latex \overrightarrow r=cos\omega ti+sin\omega tj$ where $latex \omega$ is constant, prove that. [6]
i) Velocity $latex \overrightarrow v$ of the particle is perpendicular to $latex \overrightarrow r$
ii) $latex \overrightarrow r\times\overrightarrow v=$ constant vector and.
iii) The acceleration $latex\overrightarrow a$ is directed towards the origin.

i) The possition vector is given by

$latex \begin{array}{l}\overline r=\cos wt\;i+\sin wt\;j\\\\\therefore\overline v=\frac{dr}{dt}=\left(-w\;\sin wt\right)i+\left(w\;\cos wt\right)j\\\\\therefore\overline v\cdot\overline r=\left[\left(-w\;\sin wt\right)i+\left(w\;\cos wt\right)j\right]\cdot\left[\left(\cos wt\right)i+\left(\sin wt\right)j\right]\\\\\therefore\overline v\cdot\overline r=-w\;\sin wt\;\cos wt+w\;\sin wt\;\cos wt\\\\\therefore\overline v\cdot\overline r=0\\\\\therefore\overline v\;is\;perpendicular\;to\;\overline r\end{array}$

ii) Also,

$latex \begin{array}{l}\overline r\times\overline v=\begin{vmatrix}i&j&k\\\cos wt&\sin wt&0\\w\;\sin wt&w\;\cos wt&0\end{vmatrix}\\\\\;\;\;\;\;\;\;\;=0i+0j+\left(w\;\cos^2wt+w\;\sin^2wt\right)k\\\\\;\;\;\;\;\;\;\;\;=w\left(\cos^2wt+\sin^2wt\right)k\\\\\;\;\;\;\;\;\;\;\;=wk\\\\\therefore\overline r\times\overline v=cons\tan t\;vector\end{array}$

iii) Accelaration

$latex \begin{array}{l}\overline a=\frac{d^2r}{dt^2}\\\\\;\;\;=\frac d{dt}\left[\left(-w\;\sin wt\right)i+\left(w\;\cos wt\right)j\right]\end{array}$

$latex \begin{array}{l}\overline a=\left(-w^2\;\cos wt\right)i+\left(-w^2\;\sin wt\right)j\\\\\;\;\;=-w^2\left(\cos wti+\sin wtj\right)\\\\\overline a=-w^2\;\overline r\end{array}$

b) A particle moves along the curve $latex \overline r=\left(t^3-4t\right)i+\left(t^2+4t\right)j+\left(8t^2-3t^3\right)k$ where t is the time. Find the magnitude of the tangential and normal component of its acceleration at t = 2. [6M]

$latex \overline r=\left(t^3-4t\right)i+\left(t^2+4t\right)j+\left(8t^2-3t^3\right)k$

∴ velocity,

$latex \begin{array}{l}\overline v=\frac{d\overline r}{dt}\\\\\;\;\;=\left(3t^2-4\right)i+\left(2t+4\right)j+\left(16t-9t^2\right)k\end{array}$

Accelaration

$latex \begin{array}{l}\overline a=\frac{d^2r}{dt}\\\\\;\;\;=6t\;\overline i-2j+\left(16-18t\right)k\end{array}$

at t = 2,

$latex \begin{array}{l}v=\left(3\left(2\right)^2-4\right)i+\left(2\times2+4\right)j+\left(16\times2-9\left(2\right)^2\right)k\\\\v=8i+8j-4k.\end{array}$

$latex \begin{array}{l}a=6\times2i-2j+\left(16-18\times2\right)k\\\\a=12i-2j-20k\end{array}$

unit tangent vector

$latex \begin{array}{l}\widehat t=\frac{\overline v}{\left|\overline v\right|}\\\\\;\;\;=\frac{8i+8j-4k}{\sqrt{64+64+16}}\\\\\;\;\;=\frac{8i+8j-4k}{12}\end{array}$

Tangentail component of accelaration, $latex a_t=\overline a\cdot\widehat t$

$latex \begin{array}{l}=\left(12i+2j-20k\right)\cdot\frac{8i+8j-4k}{12}\\\\=\frac{12\times8+2\times8-20\times-4}{12}\\\\=\frac{192}{12}\\\\at=16\end{array}$

consider

$latex \begin{array}{l}\widehat t\times\overline a=\begin{bmatrix}i&j&k\\\frac8{12}&\frac8{12}&\frac{-4}{12}\\12&2&-20\end{bmatrix}\\\\\;\;\;\;\;\;\;\;\;=\frac4{12}\times2\begin{bmatrix}i&j&k\\2&2&-1\\6&1&-10\end{bmatrix}\\\\\;\;\;\;\;\;\;\;\;=\frac23\left[i\left(-20+1\right)-j\left(-20+6\right)+k\left(2-12\right)\right]\\\\\widehat t\times\overline a=\frac23\left[-19i+14j-10k\right]—–\left(i\right)\end{array}$

∴ Normal component of acceleration

$latex \begin{array}{l}a_n=\left|\widehat t\times\overline a\right|\\\\\;\;\;=\left|\frac23\right.\left[-19i+14j-10k\right]\\\\\;\;\;=\frac23\sqrt{\left(-19\right)^2+14^2+\left(-10\right)^2}\\\\\;\;\;=\frac23\sqrt{361+196+100}\\\\\;\;\;=\frac23\sqrt{657}\\\\\;\;\;=\frac23\sqrt{9\times73}\\\\a_n=2\sqrt{73}\end{array}$

∴ magnitude of tangential component of accelaration at = 16 and

magnitude of normal component of accelaration $latex a_n=2\sqrt{73}$

c) Find the value of ‘n’ for which the vector field $latex r^n\overrightarrow r$ will be solenoidal. Find also whether the vector field $latex r^n\overline r$ is irrotational or not. [6M]

A vector $latex r^n\overline r$ is said to be solenoidal if Div $latex \left(r^n\;\overline r\right)=0\;where\;\overline r=xi+yi+zk$

$latex \begin{array}{l}\therefore\nabla\left(r^n\;\overline r\right)=\left(i\frac\partial{\partial x}+j\frac\partial{\partial y}+k\frac\partial{\partial z}\right)\;\left(r^nxi+r^nyj+r^nzk\right)=0\\\\\therefore\frac\partial{\partial x}r^nx+\frac\partial{\partial y}r^ny+\frac\partial{\partial y}r^nz=0\\\\\therefore r^n+xnr^{n-1}\frac{\partial r}{\partial x}+r^n+ynr^{n-1}\frac{\partial r}{\partial y}+r^n+nzr^{n-1}\frac{\partial r}{\partial z}=0\\\\\left(\because\frac d{dx}u\cdot v=u\frac{dv}{dx}+v\frac{du}{dx}\right)\\\\\therefore3r^n+\left[x\frac{\partial r}{\partial x}+y\frac{\partial r}{\partial y}+z\frac{\partial r}{\partial z}\right]nr^{n-1}=0\\\\\therefore3r^n+\left[x\cdot\frac xr+y\cdot\frac yr+z\cdot\frac zr\right]r^{n-1}n=0\\\\\left(\begin{array}{l}\because r=xi+yj+zk\\\;\;\;r^2=x^2+y^2+z^2\end{array}\right)\\\\\therefore3r^n+\frac nr+\left(x^2+y^2+z^2\right)r^{n-1}n=0\\\\3r^n+\frac nrr^nr^{n-1}=0\\\\\therefore3r^n+nr^n=0\\\\\left(3+n\right)r^n=0\\\\3+n=0\\\\\therefore n=-3\end{array}$

A vector $latex r^n\overline r$ is irrotational if $latex \nabla\times r^n\overline r=0$

$latex \therefore\nabla\times r^n\overline r=\begin{vmatrix}i&j&k\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\r^nx&r^ny&r^nz\end{vmatrix}$

$latex =i\left(\frac\partial{\partial y}\left(r^nz\right)-\frac\partial{\partial z}r^ny\right)+j\left(\frac\partial{\partial z}\left(r^nx\right)-\frac\partial{\partial x}r^nz\right)k\left(\frac\partial{\partial x}\left(r^ny\right)-\frac\partial{\partial y}r^nx\right)$

$latex =\left[nr^{n-1}z\frac{\partial r}{\partial y}-nr^{n-1}y\frac{\partial r}{\partial z}\right]i+j\left[nr^{n-1}x\frac{\partial r}{\partial z}-nr^{n-1}z\frac{\partial r}{\partial x}\right]+k\left[nr^{n-1}y\frac{\partial r}{\partial x}-nr^{n-1}x\frac{\partial r}{\partial y}\right]$

$latex =\left[nr^{n-1}z\frac yr-nr^{n-1}y\frac zr\right]i+\left[nr^{n-1}x\frac zr-nr^{n-1}z\frac xr\right]j+\left[nr^{n-1}y\frac xr-nr^{n-1}x\frac yr\right]k$

= 0

∴ rnr is irrotational

∴ The value of n=-3 and rn$latex \overline r$ is irrotational

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