Question

A particle moves in the xy plane as v = a$\widehat{i}$ + bx$\widehat{j}$ where $\widehat{i}$ and $\widehat{j}$ are the unit vectors along x and y axis. The particle starts from origin at t = 0. Find the radius of curvature of particle as a function of x.

• A. $\frac{a^{2} + b^{2}x^{2}}{ba}$
• B. $\frac{a}{b}\left\lbrack 1 + \left( \frac{bx}{a} \right)^{2} \right\rbrack^{3/2}$
• C. $\frac{b}{a}\left\lbrack 1 + \left( \frac{ax}{b} \right)^{2} \right\rbrack^{3/2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{a}{b}\left\lbrack 1 + \left( \frac{bx}{a} \right)^{2} \right\rbrack^{3/2}$

Answer 2

$\frac{dv}{dt}$ = a or x = at

$\frac{dy}{dt} = bat$ or y = $\frac{bat^{2}}{2}$ or y = $\frac{bx^{2}}{2a}$

$\frac{dy}{dx} = \frac{b}{a}x$ and $\frac{d^{2}y}{dx^{2}} = \frac{b}{a}$

R = $\frac{\left\lbrack 1 + \left( \frac{dy}{dx} \right)^{2} \right\rbrack^{3/2}}{\frac{d^{2}y}{dx^{2}}} = \frac{\left\lbrack 1 + \left( \frac{b}{a}x \right)^{2} \right\rbrack^{3/2}}{\frac{b}{a}}$=

$\frac{a}{b}\left\lbrack 1 + \left( \frac{bx}{a} \right)^{2} \right\rbrack^{3/2}$