This is exercise 4 page 202 from Spivak's book on mechanics. Two particles of masses $m_1,m_2$ are attached to a lever of negligible weight, and are at distances $d_1,d_2$ from the fulcrum, like this:
Suppose we start the lever making an angle $\phi_0$ with the horizontal, and let $T$ be the time it takes for the lever to reach a horizontal position, so that the first particle has fallen the distance $d_1 \sin(\phi_0)$. We are going to consider $d_2$ to be fixed, while $d_1$ can be varied, by moving the first particle along the lever. Letting $D=d_1/d_2$, show that $T$ satisfies
$$T^2=\text{constant}\times D \frac{Dm_1+m_2}{Dm_1-m_2}$$
If anyone is aware of this problem your help will be greatly appreciated! At the moment, I doubt the statement is correct, for the following reason: I used d'Alembert's principle to derive the equations of motion, using as variables the distance $r$ of the first particle from the fulcrum and the angle $\phi$ that the lever makes with the horizontal: $$r''=r(\phi')^2-g\sin(\phi)$$ and $$(m_1r^2+m_2d_2^2)\phi''=-g\cos(\phi)(m_1r-m_2d_2)-2m_1rr'\phi'$$ To understand what these equations say, for $r''$ the first term on the RHS is the centrifugal force, and the second is the effect of gravity. For $\phi''$, the equation can be rewritten as $$((m_1r^2+m_2d_2^2)\phi')' = -g\cos(\phi)(m_1r-m_2d_2)$$ where the LHS is the derivative of the angular momentum and on the RHS the total torque. Now, because the equations are complex to solve (even after using that work=change of kinetic energy), I tried to understand the problem by looking at the case $\phi_0 \approx 0$. This way, it will take very little time to reach the horizontal, and in the meantime we can approximate $$\phi(t)\approx \phi_0 +\frac{1}{2}\phi''(0)t^2$$ Now, $\phi''(0)\approx-g\frac{m_1d_1-m_2d_2}{m_1d_1^2+m_2d_2^2}$. Since $\phi(T)=0$, we get $\phi_0 +\frac{1}{2}\phi''(0)T^2 \approx 0$ and so $$T^2\approx \frac{2\phi_0}{g}\frac{m_1d_1^2+m_2d_2^2}{m_1d_1-m_2d_2}=\text{constant}\times d_2\frac{D^2m_1+m_2}{Dm_1-m_2}$$ and this is different than what Spivak says.
With
$$ \cases{ p_1 = d_1(\cos\theta,\sin\theta)\\ p_2 =-d_2(\cos\theta,\sin\theta) } $$
we have
$$ \mathcal{L}(d_1,\theta) = \frac 12(m_1\dot p_1^2+m_2\dot p_2^2)-g(m_1p_1+m_2p_2)\cdot \hat e_y $$
where $\hat e_y = (0,1)$. Developing the movement equations we have
$$ \frac{d}{dt}\nabla_{\dot d_1,\dot\theta}\mathcal{L}-\nabla_{d_1,\theta}\mathcal{L}=\cases{\ddot d_1 -d_1\dot\theta^2+g\sin\theta = 0\\ (m_1d_1^2+m_2d_2^2)\ddot\theta + 2m_1d_1\dot d_1\dot\theta+g\cos\theta(m_1d_1-m_2d_2)=0} $$
From the last equation, assuming $\theta \approx \theta_0+\dot\theta_0 t + \frac{\ddot\theta_0}{2}t^2$ with $\dot\theta_0 = 0$ and $\ddot \theta_0 = -c_0$ we have
$$ -(m_1d_1^2+m_2d_2^2)\frac{c_0}{2}t^2 +g\cos\theta_0(m_1d_1-m_2d_2)=0 $$
and thus
$$ t^2\approx \frac{2g\cos\theta_0}{c_0}\left(\frac{m_1d_1-m_2d_2}{m_1d_1^2+m_2d_2^2}\right) $$