Question

A particle moves along the plane trajectory y(x) with constant speed v. Find the radius of curvature of the trajectory at the point x = 0 if the trajectory has the form of a parabola $y=a{x}^2$ where 'a' is a positive constant.

Answer 1

If the equation of the trajectory of a particle is given we can find the radius of trajectory of the instantaneous circle by using the formula
$R=\frac{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3/2}}{\left|\frac{d^{2}y}{d{x}^2}\right|}$
As ; $y=a{x}^2\Rightarrow \frac{dy}{dx}=2ax=0(atx=0)\frac{d^{2}y}{d{x}^2}=2a$
Now radius of trajectory is given by
$R=\frac{{[1+0]}^{3/2}}{2a}=\frac{1}{2a}$
This problem can also be solved by using the formula : $R=\frac{v^2}{a_{\perp }}.y=a{x}^2,$
differentiate with respect to time $\frac{dy}{dt}=2ax\frac{dx}{dt}$............(1)
at $x=0,v_y=\frac{dy}{dt}=0hence{v}_x=v$
since $v_x$ is constant, $a_x=0$
Now, differentiate (1) with respect to time $\frac{d^{2}y}{d{t}^2}=2ax\frac{d^{2}x}{d{t}^2}+2a{\left(\frac{dx}{dt}\right)}^2$
at $x=0,v_x=v$
$\therefore$ net acceleration, $a=a_y=2a{v}^2(since{a}_x=0)$
this acceleration is perpendicular to velocity $\left(v_x\right)$
Hence it is equal to centripetal acceleration
$R=\frac{v^2}{a_{\perp }}=\frac{v^2}{2a{v}^2}=\frac{1}{2a}$