Question

A particle moves along the plane trajectory $y\left(x\right)$ with velocity $v$ whose modulus is constant. Find the radius of curvature at origin if the trajectory has the form of a parabola $y=a{x}^2$.

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

Answer 1

The correct option is D $R=\frac{1}{2}a$

We have $y=a{x}^2$

So, $\frac{dy}{dt}=2ax\frac{dx}{dt}$
or, $v_y=2ax{v}_x$
or, $a_y=2ax{a}_x+2a{v}_x^2$

At $x=0$, we have:
$v_y=0$ and $v_x=v$
$\Rightarrow$ $a_y=2a{v}^2$

As speed is constant, the tangential acceleration would be zero.

Thus $a_x=\frac{dv}{dt}=0$
$\Rightarrow$ $a_{net}=2a{v}^2$

Now radius of curvature:
$R=\frac{v^2}{a_{net}}=\frac{v^2}{2a{v}^2}$
or, $R=\frac{1}{2a}$