Rectangle $ABCD$ is tilted such that its base $AB$ makes an angle of $\theta$ with the horizontal floor. It has its vertex $A$ in contact with the horizontal floor, and the rectangle is released from this position, to fall freely and the vertex at $A$ is free to slide along the floor.
What will be the rectangle's final linear and angular velocity when its base hits the floor?
The rectangle has dimensions: $\overline{AB} = \ell$ and $\overline{BC} = w $.
My initial thought:
The solution, I think, can be obtained from the Euler-Lagrange equation of motion.
First, I define my state variables as follows: firstly, we have the linear distance of point $A$ from a fixed origin. Let's call this $x$. Then we have the tilt angle $q$ of the base $AB$ from the $x$ axis. We can take $x(0) = 0$ and $q(0) = \theta$.
The position of the center of mass of the rectangle is $$ P = (x, 0) + ( \ell \cos(q) - w \sin(q), \ell \sin(q) + w \cos(q) ). $$ Hence, the linear velocity is $$ \dot{P} = (\dot{x}, 0) + \dot{q} ( - \ell \sin(q) - w \cos(q) , \ell \cos(q) - w \sin(q) ). $$ The kinetic energy due to $\dot{P}$ is $$ \dfrac{1}{2} m \dot{P}^2 = \dfrac{1}{2} m \Big( \dot{x}^2 + \dot{q}^2 ( \ell^2 + w^2 ) + 2 \dot{x} \dot{q} \big( - \ell (q) - w \cos(q) \big) \Big). $$ And the kinetic energy due to rotation is $$ \dfrac{1}{2} I \dot{q}^2. $$ where $I$ is the moment of ineratia about the corner $A$ and is given by $$ I = \dfrac{1}{3} (\ell^2 + w^2). $$ Hence, the overall kinetic energy is $$ T = \dfrac{1}{2} m \dot{x}^2 + \dfrac{2}{3} m \dot{q}^2 + m \dot{x} \dot{q} ( - \ell \sin(q) - w \cos(q) ) $$ The potential energy is $$ V = \dfrac{1}{2} m g \big( \ell \sin(q) + w \cos(q) \big). $$ And as usual we define the Lagrangian function $L$ as follows $$ L = T - V = \dfrac{1}{2} m \dot{x}^2 + \dfrac{2}{3} m (\ell^2 + w^2) \dot{q}^2 + \dot{x} \dot{q} ( - \ell \sin(q) - w \cos(q) ) - \dfrac{1}{2} m g ( \ell \sin(q) + w \cos(q) ). $$ And the Euler-Lagrange equations of motion are $$ \dfrac{d}{dt} \dfrac{\partial L}{\partial \dot{q}_i } = \dfrac{\partial L }{ q_i}. $$ where $q_1 = x, q_2 = q $. Applying this equation to our $L$, give the following two equations $$ \begin{cases} \ddot{x} + \ddot{q} \big( - \ell \sin(q) - w \cos(q) \big) + \dot{q}^2 \big( - \ell \cos(q) + w \sin(q) \big) = 0\\ \\ \dfrac{4}{3} (\ell^2 + w^2) \ddot{q} + \ddot{x} \big(-\ell \sin(q) - w \cos(q) \big) + \dot{x} \dot{q} \big( - \ell \cos(q) + w \sin(q) \big) = - \dfrac{1}{2} g \big( \ell \cos(q) - w \sin(q) \big). \end{cases} $$
Finally, if I adopt the suggestion in the comment below by mjqxxxx of changing the varible $q$ to $\phi = q + \tan^{-1}\left( \dfrac{w}{l} \right) $, and defining $r = \dfrac{1}{2}\sqrt{\ell^2 + w^2}$, then
$ \cos \phi =( \cos q \ell - \sin q w ) / (2 r ) $
$ \sin \phi = ( \sin q \ell + \cos q w ) / ( 2 r ) $
And $ \dot{q} = \dot{\phi} $ and $ \ddot{q} = \ddot{\phi} $
With this the equations above become
$$ \begin{cases} \ddot{x} - 2 r \ddot{q} \big( \sin \phi \big) - 2 r \dot{q}^2 \cos \phi = 0\\ \\ \dfrac{16}{3} r^2 \ddot{q} -2 r \ddot{x} \sin \phi - 2 r \dot{x} \dot{q} \cos \phi = - r g \cos \phi \end{cases} $$
Substituting for $\ddot{x}$ from the first into the second gives
$\dfrac{16}{3} r^2 \ddot{q} -4 r^2 \bigg( \ddot{q} \big( \sin \phi \big) + \dot{q}^2 \cos \phi \bigg) \sin \phi - 2 r \dot{x} \dot{\phi} \cos \phi = - r g \cos \phi$
We also know that due of absence of horiztonal forces, the horizontal coordinate of the center of mass of the rectangle stays constant, but this is given by $ x + r \cos \phi $, therefore, $ \dot{x} - r \dot{\phi} \sin \phi = 0 $. Substituting this, we get an equation in $\phi$ and its two time derivatives:
$\dfrac{16}{3} r^2 \ddot{\phi} -4 r^2 \bigg( \ddot{\phi} \big( \sin \phi \big) + \dot{\phi}^2 \cos \phi \bigg) \sin \phi - 2 r^2 \sin \phi \dot{\phi}^2 \cos \phi = - r g \cos \phi$
Divide by $r^2$ and simplify
$ ( \dfrac{4}{3} + 4 \cos^2 \phi ) \ddot{\phi} - 6 \dot{\phi}^2 \sin \phi \cos \phi = - \left(g/ r \right) \cos \phi $
Hence,
$ \ddot{\phi} = \dfrac{ - 3 (g/r) \cos \phi + 18 \dot{\phi}^2 \sin \phi \cos \phi }{ 4 + 12 \cos^2 \phi } $
I have two questions now:
- Are these equations correct?
- And second, are there closed-form solutions for them?
This may be a case where the Euler-Lagrange machinery is unnecessarily abstract. Call the center of the rectangle $O$; let $\phi$ be the angle between the positive $x$-axis and $\overline{AO}$; and let $r=\frac{1}{2}\sqrt{w^2+\ell^2}$ be the length of $\overline{AO}$. The center of mass is at height $h=r \sin\phi$. There is an upward force $F$ acting at $A$, and the downward force of gravity $mg$ acting at $O$, so the net force is $F-mg$, and $$ F-mg=m\ddot{h}=m\frac{d^2}{dt^2}(r\sin\phi)=mr\left(\ddot{\phi}\cos\phi - (\dot{\phi})^2\sin\phi\right). $$ Moreover, the torque relative to $O$ is $\tau=-Fr\sin(\pi/2-\phi)=-Fr\cos\phi$, so $$ -Fr\cos\phi=I\ddot{\phi}=\frac{1}{3}mr^2\ddot{\phi}. $$ Putting these two expressions for $F$ together, we find the equation of motion for $\phi$ to be $$ \ddot{\phi}=\frac{(\dot{\phi})^2\sin\phi\cos\phi - (g/r)\cos\phi}{\cos^2\phi + 1/3}. $$ Note that $\ddot\phi$ is initially negative, as expected, since the block is rotating clockwise and $\phi$ will decrease. Also, this equation of motion appears to depend on $g/r$, but it really doesn't... choosing the unit of time to be $\sqrt{r/g}$ makes the equation dimensionless and parameter-free. Two things can happen as $\phi$ decreases at an ever-increasing rate. If $\phi$ reaches $\tan^{-1}(w/\ell)$, the bottom of the block hits the ground. Or, if $F$ becomes zero first, the corner of the block will leave the ground; this occurs when $(\dot{\phi})^2 \ge g/(r\sin\phi)$. It turns out that the block doesn't leave the ground if we start at rest (though it could if the block were given an initial angular momentum).