Question

A particle is moving in the $x-y$ plane with uniform speed $v$. The equation of trajectory for the particle is represented by the parabola $y=a{x}^2,$ where $a$ is a $+ve$ constant. Find the radius of curvature of trajectory at the point $x=0$.

• A. $\frac{a}{2}$
• B. $\frac{1}{2a}$
• C. $\frac{4}{a}$
• D. $\frac{3}{2}a$

Answer 1

The correct option is B $\frac{1}{2a}$

Point $x=0$ will represent position at the origin $(0,0)$, since $y=0$ at $x=0$ from the equation of trajectory given by $y=a{x}^2$
Differentiating w.r.t time gives:
$\frac{dy}{dt}=2ax\frac{dx}{dt}...\left(i\right)$
$\Rightarrow v_y=2ax{v}_x...\left(ii\right)$

Putting, $x=0$ in Eq. $\left(i\right)$ gives $v_y=0$, hence it will have only velocity in $x-$ direction i.e $v_x$ at $x=0$. The motion parameters have been shown in the figure given below:


$\because$ Particle is moving at uniform speed $v$
$\Rightarrow v_x=v...\left(iii\right)$
Again differentiating Eq. $\left(i\right)$ w.r.t time:
$\frac{d^{2}y}{d{t}^2}=2a[\frac{dx}{dt}.\frac{dx}{dt}+x.\frac{d^{2}x}{d{t}^2}]$
$\Rightarrow a_y=2a[{\left(\frac{dx}{dt}\right)}^2+x.\frac{d^{2}x}{d{t}^2}]$
At point $x=0$,
$a_y=2a[{\left(\frac{dx}{dt}\right)}^2+0]=2a{v}_x^2$
From Eq. $\left(iii\right)$, we get
$a_y=2a{v}^2$

As shown in figure at $(x=0,y=0)$, net acceleration of the particle is $a_y$ which is $\perp$ to net velocity ($v_x$).
$\therefore a_y=\frac{(v_x{)}^2}{R}$ where $R$ is the radius of curvature.
$\Rightarrow 2a{v}^2=\frac{v^2}{R}$
$\therefore R=\frac{1}{2a}$ is the radius of curvature of the trajectory at $x=0$.