Two particles of mass m and 2m are hung from the ends of a uniform chain of mass m and length 2a. The chain is hung from a smooth peg such that its length on the two sides are equal and it is then allowed to fall. Show that the chain leaves the peg after a time $ \ \sqrt{\frac{4a}{g}} \ \ln(2+ \sqrt3) \ $.
How do I show this?
On
The total mass of the system is $4m$. From Newton's Law: $F(x) = 4m a = 4 m x^{\prime\prime}(t) = g \left( m + \frac{2 x m}{2 a} \right)$, with $x(0) = x^\prime (0) = 0$ and acceleration $a(t) = x^{\prime\prime} (t)$.
Solve to find:
$$x(t) = 2 a \sinh ^2\left(\frac{\sqrt{g} t}{4 \sqrt{a}}\right)$$
Solve for when $x(t) = a$.
Since you mentioned working with a force diagram, I'm not sure whether you have learned to use an energy-analysis yet (though you will likely do so soon, if you haven't already).
The "smooth" peg is taken to be frictionless, so this problem only deals with gravitational potential energy and kinetic energy of the two masses and the "massive" (as opposed to massless) chain. Since the chain has mass $ \ m \ $ and length $ \ 2a \ , $ its linear mass density is $ \ \lambda \ = \ \frac{m}{2a} \ . $ Initially then, there is a mass $ \ m \ $ and a chain mass of $ \ \lambda · a \ $ to one side of the peg and, to the other side, a mass $ \ 2m \ $ and a chain mass also of $ \ \lambda · a \ . $
After the masses are released and the masses and chain have moved by a distance $ \ x \ , \ $ the lighter mass has ascended , the heavier mass has descended, and the length of chain on each side of the peg has likewise changed. The changes in gravitational potential energy of the masses are $ \ +mg · x \ $ and $ \ -2mg · x \ . $ Since the chain has constant density, the now shorter length of chain can be treated as having length $ \ a - x \ $ with its center of mass lifted by $ \ \frac{x}{2} \ $ , while the now greater length of chain, $ \ a + x \ , $ has its center of mass lowered by the same amount. So the change in gravitational potential energy of the system is
$$ \left[ \underbrace{\ +mgx \ + \ \lambda · (a-x)g · \left( \frac{x}{2} \right)}_{\text{ascending mass}} \ \right] \ + \ \left[ \underbrace{\ -2mgx \ - \ \lambda · (a+x)g · \left( \frac{x}{2} \right)}_{\text{descending mass}} \ \right] $$ $$ = \ \ -mgx \ - \ 2 · \lambda g · x · \left( \frac{x}{2} \right) \ \ = \ \ -mgx \ - \ \frac{mg}{2a} · x^2 \ \ . $$
Because the system is frictionless, and has total mass $ \ 4m \ , $ the change in its kinetic energy is exactly the negative of this, allowing us to write
$$ \ \frac{1}{2} · (4m) · v^2 \ = \ 2m · \left( \frac{dx}{dt} \right)^2 \ \ = \ \ mg · \left( x \ + \ \frac{1}{2a} · x^2 \right) $$ $$ \Rightarrow \ \ \frac{dx}{dt} \ \ = \ \ \sqrt{ \frac{g}{2} · \left( x \ + \ \frac{1}{2a} · x^2 \right) } \ \ , $$
giving us a separable differential equation which we integrate in time from $ \ 0 \ $ to $ \ T \ $ and in distance from $ \ 0 \ $ to $ \ a \ $ thus,
$$ \int_0^a \ \frac{dx}{\sqrt{ \left( x \ + \ \frac{1}{2a} · x^2 \right) }} \ \ = \ \ \int_0^T \sqrt{\frac{g}{2}} \ \ dt \ \ , $$
with $ \ T \ $ being the time required for the ascending mass to reach the peg, and where $ \ x(0) \ = \ 0 \ $ and $ \ v(0) \ = \ 0 \ \ . $
The integration leaves us with
$$ 2 \ \sqrt{2a} \ \sinh^{-1} \left( \sqrt \frac{x}{2a} \right) \left\vert_0^a \right . \ = \ \sqrt{\frac{g}{2}} · T \ \ \ \text{or} \ \ \ T \ = \ 4 \ \sqrt{\frac{a}{g}} \ \ln \left[ \ \left( \sqrt \frac{x}{2a} \right) \ + \ \sqrt{1 + \frac{x}{2a}} \ \right] \left\vert_0^a \right . \ \ . $$
This latter expression produces
$$ T \ = \ 4 \ \sqrt{\frac{a}{g}} \ \ln \left[ \ \left( \sqrt \frac{a}{2a} \right) \ + \ \sqrt{1 + \frac{a}{2a}} \ \right] \ \ = \ \ 4 \ \sqrt{\frac{a}{g}} \ \ln \left[ \ \sqrt \frac{1}{2} \ + \ \sqrt{\frac{3}{2}} \ \right] $$ $$ = \ \ 2 \ \sqrt{\frac{a}{g}} \ \ln \left[ \ \left( \ \sqrt \frac{1}{2} \ + \ \sqrt{\frac{3}{2}} \ \right)^2 \ \right] \ \ = \ \ 2 \ \sqrt{\frac{a}{g}} \ \ln \left( \ 2 \ + \ \sqrt{ 3 } \ \right) \ \ . $$