Question

A point moves in the plane so that its tangential acceleration ${\omega }_{\tau }=a$, and its normal acceleration ${\omega }_n=b{t}^4$, where $a$ and $b$ are positive constants, and $t$ is time. At the moment $t=0$ the point was at rest. Find how the curvature radius $R$ of the point's trajectory and the total acceleration $\omega$ depend on the distance covered $s$.

• A. $R=\frac{a^3}{2bs}$
• B. $\omega =a\sqrt{1+{\left(\frac{4b{s}^2}{a^3}\right)}^2}$
• C. $R=\frac{a^3}{4bs}$
• D. $\omega =a\sqrt{1+{\left(\frac{2b{s}^2}{a^3}\right)}^2}$

Answer 1

Answer: (A), (B)

The correct options are
A $R=\frac{a^3}{2bs}$
B $\omega =a\sqrt{1+{\left(\frac{4b{s}^2}{a^3}\right)}^2}$

As ${\omega }_t=a$ and at $t=0$, the point is at rest.
So, $v\left(t\right)$ and $s\left(t\right)$ are, $v=at$ and $s=\frac{1}{2}a{t}^2$ .................................(1)
Let $R$ be the curvature radius, then
${\omega }_n=\frac{v^2}{R}=\frac{a^2{t}^2}{R}=\frac{2as}{R}$ (using 1)
But according to the problem: ${\omega }_n=b{t}^4$
So, $b{t}^4=\frac{a^2{t}^2}{R}$

or, $R=\frac{a^2}{b{t}^2}=\frac{a^3}{2bs}$ (using 1) .........................(2)
Therefore, $\omega =\sqrt{{\omega }_t^2+{\omega }_n^2}=\sqrt{a^2+{\left(\frac{2as}{R}\right)}^2}=\sqrt{a^2+{\left(\frac{4b{s}^2}{a^2}\right)}^2}$ (using 2)
Hence, $\omega =a\sqrt{1+{\left(\frac{4b{s}^2}{a^3}\right)}^2}$