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UPSC 2016 Maths Optional Paper 1 Q8c — Step-by-Step Solution

15 marks · Section B

Rectilinear motion under variable force · Dynamics & Statics · asked 3× in 14 yrs · Read the full method →

Question

A particle moves in a straight line. Its acceleration is directed towards a fixed point $O$ in the line and is always equal to $\mu\left(\dfrac{a^5}{x^2}\right)^{1/3}$ when it is at a distance $x$ from $O$. If it starts from rest at a distance $a$ from $O$, then find the time, the particle will arrive at $O$.

Technique

$v\,dv/dx=$ acceleration to get $v(x)$; then $T=\int_0^a dx/v$; substitution $x=a s^3$ turns it into the Beta integral $\int_0^1 s^2(1-s)^{-1/2}ds=\tfrac{16}{15}$.

Solution

The acceleration magnitude is $\mu\big(a^5/x^2\big)^{1/3}=\mu\,a^{5/3}x^{-2/3}$, directed toward $O$ (i.e. in the $-x$ direction). With the particle at distance $x>0$,

Step 1 — First integral (energy / speed)

Use $\ddot x=v\dfrac{dv}{dx}$:

Integrate, with $v=0$ at $x=a$:

At $x=a$: $0=-3\mu a^{5/3}a^{1/3}+\text{const}\Rightarrow\text{const}=3\mu a^{2}$. Hence

Step 2 — Set up the time integral

Moving toward $O$, $x$ decreases so $v=\dfrac{dx}{dt}<0$; take $\dfrac{dx}{dt}=-\sqrt{6\mu a^{5/3}\big(a^{1/3}-x^{1/3}\big)}$. The time from $x=a$ to $x=0$ is

Step 3 — Substitution

Let $x=a\,s^3$ (so $x^{1/3}=a^{1/3}s$), $dx=3a\,s^2\,ds$; $s:0\to1$. Then $a^{1/3}-x^{1/3}=a^{1/3}(1-s)$ and

So

Step 4 — Evaluate $\displaystyle\int_0^1\frac{s^2}{\sqrt{1-s}}\,ds$

This is a Beta integral: with $u=1-s$,

Therefore

Rationalising, $\dfrac{16}{5\sqrt6}=\dfrac{16\sqrt6}{30}=\dfrac{8\sqrt6}{15}$, so

Answer