Find the acceleration $ω$ of the particle at the point $x = 0$ if the trajectory has the form of an ellipse ${(\frac{x}{a})}^{2} + {(\frac{y}{b})}^{2} = 1$ ; $a$ and $b$ are constants here.

ω = \frac{b v^{2}}{2 a^{2}}

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ω = \frac{3 b v^{2}}{2 a^{2}}

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ω = \frac{3 b v^{2}}{a^{2}}

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ω = \frac{b v^{2}}{a^{2}}

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Solution

The correct option is C $ω = \frac{b v^{2}}{a^{2}}$
${(\frac{x}{a})}^{2} + {(\frac{y}{b})}^{2} = 1 . . . . (1)$
differentiating with respect to $x$ .
$\frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{d y}{d x} = 0$
$\frac{d y}{d x} = - \frac{x b^{2}}{y a^{2}} . . . . . . (2)$
Again differentiating with respect to $x$ .
$\frac{d^{2} y}{d x^{2}} = - \frac{b^{2}}{a^{2}} [x (- \frac{1}{y^{2}}) \frac{d y}{d x} + \frac{1}{y}]$
$\frac{d^{2} y}{d x^{2}} = - \frac{b^{2}}{a^{2}} [- \frac{x^{2} b^{2}}{y^{3} a^{2}} + \frac{1}{y}]$ using (2)
Radius of curvature , $R = \frac{{[1 + (\frac{d y}{d x})^{2}]}^{3 / 2}}{\frac{d^{2} y}{d x^{2}}}$
$R = \frac{{[1 + (\frac{x^{2} b^{2}}{y^{2} a^{2}})^{2}]}^{3 / 2}}{- \frac{b^{2}}{a^{2}} [- \frac{x^{2} b^{2}}{y^{3} a^{2}} + \frac{1}{y}]}$
At $x = 0, R = - \frac{a^{2}}{b^{2}} y$
using (1) for $x = 0, y = b o r - b$
thus , $R = \frac{a^{2}}{b}$
$∴$ Acceleration , $ω = \frac{v^{2}}{R} = \frac{b v^{2}}{a^{2}}$

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Find the acceleration ω of the particle at the point x=0 if the trajectory has the form of an ellipse (xa)2+(yb)2=1; a and b are constants here.

Find the acceleration $ω$ of the particle at the point $x = 0$ if the trajectory has the form of an ellipse ${(\frac{x}{a})}^{2} + {(\frac{y}{b})}^{2} = 1$ ; $a$ and $b$ are constants here.