# Block sliding down a table

Disclaimer: I came up with the following problem on my own, so I don't know if there is a solution that makes sense our if my approach to solving it is correct... Any help is appreciated.

Consider the following situation: A rectangle of length $$l=0.1$$ and width $$b=0.02$$ is sliding along a table with some initial velocity. I'd like to calculate the trajectory and the time until it loses contact to the table.

My idea here was to use Lagrangian mechanics. To simplify the explanation I drew the following sketch for some time $$t>0$$. The red dot should represent the center of mass and $$s$$ the distance between the center of mass and the point of rotation (lets set the center of our coordinate system there). We can now write the center of mass as

$$\vec{r}(t) = \begin{pmatrix}s(t)\cos\theta(t) + \frac{b}{2}\sin\theta(t)\\ -s(t)\sin\theta(t) +\frac{b}{2}\cos\theta(t)\end{pmatrix}.$$

%With this we are able to write down the Lagrangian in the form $$L= T-V =\frac{1}{2}m\dot{\vec{r}}^2+\frac{1}{2}J\dot\theta^2+mgs\sin\theta-\frac{mgb}{2}\cos\theta.$$

We further find \begin{align*} \frac{1}{2}m\dot{\vec{r}}^2 &= \frac{1}{8} m (4 {\dot s}^2 +4 b \dot s\dot\theta + (b^2 + 4 s^2) \dot\theta^2)\\ \frac{1}{2}J\dot\theta^2 &= \frac{1}{2} (\frac{1}{12} m (l^2 + b^2) + m s^2) \dot\theta^2 \end{align*}, where I used the parallel axis theorem. All of this together gives $$L = \frac{1}{24}(l^2\dot\theta^2+4b^2\dot\theta^2+24s\dot\theta^2+12\dot s^2+ 12 b\dot s \dot\theta+24gs\sin\theta -12bg\cos\theta).$$ The equations of motion are then given by \begin{align*} \dot\theta^2 +g \sin\theta &=\ddot s + \frac{1}{2}b\ddot\theta\\ \left(\frac{1}{12}l^2+\frac{1}{3}b^2\right)\ddot\theta+\frac{1}{2}b\ddot s &= \frac{1}{2}bg\sin\theta-2\dot s\dot\theta-gs\cos\theta. \end{align*}

The condition for the rectangle to lose contact to the table is given by $$[\ddot{\vec{r}}]_x=0 \Longleftrightarrow \frac{d^2}{dt^2}(-s(t)\sin\theta(t) +\frac{b}{2}\cos\theta(t)) =0$$ This equation should in principle be enough to determine the time $$t^*$$ when the rectangle leaves the table.

Let us assume that we also have the following initial conditions $$s(0)=0.01,\quad \dot s(0)= 0.3,\quad u(0)= 0,\quad \dot u(0)= 0.$$ Since you can't solve this analytically I tried it numerically (Mathematica code) and got the following plots I then tried to find $$t^*$$ using my approximate solutions for $$s$$ and $$\theta$$ but if you plot $$[\ddot{\vec{r}}]_x$$ against $$t$$ you get

which clearly doesn't have a zero at all...

I checked the Mathematica code multiple times and also checked for calculation errors up to arriving at the ODE's, but I can't find an error which leads me to believe that I have a more fundamental problem in my approach. Do you see something obvious (or not so obvious) that I'm doing wrong?

• You get $s(t)>100 [m]$ impossible. I think that the if $s(t) >l/2$ the rod lose contact.
– Eli
Apr 22 '19 at 22:04
• @Eli It should lose contact a lot sooner... Just take a book, apply some force (such that its horizontal velocity isn't zero) and you'll see that the end of the book probably won't touch the rotating point.
– Sito
Apr 22 '19 at 22:07
• I understand you came up with this problem on your own, but I still feel like it fits with the "homework-and-exercises" tag, since it is essentially an exercise in Lagrangian mechanics and plotting results from such work. It is not meant to be a put-down to the question though. It is an interesting problem Apr 23 '19 at 17:54
• I haven't looked at your equations, but your graph for $\theta$ over time seems very odd. I'd expect it to accelerate, not decelerate. Apr 23 '19 at 17:55
• I am confused by your notation. It looks like the center of mass is just described by the $s$ vector if you put your origin at the corner. What is this $b$ term for? Apr 23 '19 at 17:57

First of all, I just want to point an error in the ODE's of the original post. I can tell just by looking at them that they are both wrong, because they do not make sense dimensionally. If we look at the first one: \begin{align*} \dot\theta^2 +g \sin\theta &=\ddot s + \frac{1}{2}b\ddot\theta\ \end{align*}

Everything has dimension of acceleration in this equation, except $$\dot \theta^2$$, so you probably forgot an $$s$$ or a $$b$$ along the way. Note that my solution also get rid of the parallel axis theorem, which I will explain why.

Here's my solution. I use the same coordinates as the original post. $$\vec{r}(t) = \begin{pmatrix}s(t)\cos\theta(t) + \frac{b}{2}\sin\theta(t)\\ -s(t)\sin\theta(t) +\frac{b}{2}\cos\theta(t)\end{pmatrix}$$

Then: \begin{align*} \frac{1}{2}m\dot{\vec{r}}^2 &= \frac{1}{8} m (4 {\dot s}^2 +4 b \dot s\dot\theta + (b^2 + 4 s^2) \dot\theta^2)\\ \frac{1}{2}J\dot\theta^2 &= \frac{1}{2} (\frac{1}{12} m (l^2 + b^2) \dot\theta^2 \end{align*}

Here's the key thing: You do not use the parrallel axis theorem, because then you are "overcounting" the motion of your center of mass. We only want the energy coming from the rotation of the body around its center of mass. How do we find it? Well, we know that instantaneously the center of mass rotates with angular velocity $$\dot\theta$$ around the corner of the plane. But from the perspective of the center of mass, the corner (i.e. the block) rotates with angular velocity $$\dot\theta$$ around the center of mass. Therefore the body rotates around its center of mass with angular velocity $$\dot\theta$$, so the rotationnal energy term is simply $$\frac{1}{2}J\dot\theta^2 = \frac{1}{2} (\frac{1}{12} m (l^2 + b^2) \dot\theta^2$$, where $$J$$ is the usual moment of inertia of a block around its CM.

The Lagrangian is: $$L = \frac{1}{2}m\dot{\vec{r}}^2 + \frac{1}{2}J\dot\theta^2 - mg(-s\sin\theta +\frac{b}{2}\cos\theta)$$.

The ODE's I get are: \begin{align*} s\dot\theta^2 +g \sin\theta &=\ddot s + \frac{1}{2}b\ddot\theta\\ \left(\frac{1}{12}l^2+\frac{1}{3}b^2+s^2\right)\ddot\theta+\frac{1}{2}b\ddot s + 2s\dot s\dot \theta &= \frac{1}{2} g b \sin \theta + gs\cos \theta . \end{align*}

The block looses contact when $$\ddot \theta =0$$, since the body will simply keep rotating about its CM with whatever last angular velocity it had when it left the plane. Using the initial conditions given in the originl post, we get the following plot for $$\ddot \theta$$:

The block looses contact at about 0.095 s. This result does make intuitive sense when looking at the intial velocity and at the length of the block.

• I have the same equations of motions, but I got different looses contact time?
– Eli
Apr 27 '19 at 20:30
• The block looses contact when the contact force is zero?
– Eli
Apr 27 '19 at 20:45
• @Eli Just checked Shamaz 's calculation and I get the same time as he with the condition of the force in $x$-direction being zero.
– Sito
Apr 27 '19 at 22:11

I think that this is the solution:

I put a body fixed coordinate system at the center of mass of the rechtangle , so the position vector is: $$\vec{r}=\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \\ \end{pmatrix} \begin{pmatrix} -s\\ b/2\\ \end{pmatrix}=\left[ \begin {array}{c} -\cos \left( \theta \right) s-1/2\,\sin \left( \theta \right) b\\ -\sin \left( \theta \right) s+1/2\,\cos \left( \theta \right) b\end {array} \right]$$ The kinetic energy is: $$T=\frac{1}{2}\,m\,\vec{\dot{r}}^T\,\vec{\dot{r}}=\left( 1/2\,m{s}^{2}+1/8\,m{b}^{2}+1/2\,J \right) {\dot{\theta}}^{2}+1/2 \,m{\it \dot{s}}\,\dot{\theta}\,b+1/2\,{{\it \dot{s}}}^{2}m$$ where $$J$$ is the Inertia of the rectangle at the center of mass.

The potential energy is: $$V=mg \left( -\sin \left( \theta \right) s+1/2\,\cos \left( \theta \right) b \right)$$ The equations of motion: $${\frac {d^{2}}{d{\tau}^{2}}}s \left( \tau \right) ={\frac {mbs\theta p \,{\it sp}}{m{s}^{2}+J}}+1/4\,{\frac { \left( m{b}^{2}s+4\,Js+4\,m{s}^ {3} \right) {\theta p}^{2}}{m{s}^{2}+J}}+1/4\,{\frac {4\,Jg\sin \left( \theta \right) +4\,m{s}^{2}g\sin \left( \theta \right) -2\,mbg \cos \left( \theta \right) s}{m{s}^{2}+J}}$$ and $${\frac {d^{2}}{d{\tau}^{2}}}\theta \left( \tau \right) =-2\,{\frac {ms \theta p\,{\it sp}}{m{s}^{2}+J}}-1/2\,{\frac {mbs{\theta p}^{2}}{m{s}^ {2}+J}}+{\frac {mgs\cos \left( \theta \right) }{m{s}^{2}+J}}$$ where $$sp=\dot{s}$$ and $$\theta p=\dot{\theta}$$

Simulation:

we solve the equations of motion numerically with $$J=1/12\,m(l^2+b^2)$$ .

Because the constraint force between the table and the rectangle is proportional to the acceleration of the rectangle in space coordinate system, we can check the condition where is no contact between the rectangle and table and find the time where the contact froce equal zero.

$$F_x\propto\,\vec{\ddot{r}}_x=0$$

Where $$F_x$$ is the contact force toward the x space axis.

result $$\vec{\ddot{r}}_x$$ over t

so for $$t=0.114$$[s] there is no contact between the rectangle and the table.

• Once again, can you explain to me why you are ignoring the axis around which the rectangle rotates? The moment of inertia changes according to the axis of rotation and if I‘m reading your equations correctly, you are assuming a rotation around the center of mass, which is not what is happening...
– Sito
Apr 24 '19 at 10:25
• no t don't think so, the coordinate is body fix at the center of mass.I add a figure
– Eli
Apr 24 '19 at 10:29
• Right now I don‘t see it, so I‘ll need to think about it later... Probably my first question would be what exactly $r(t)$ describes in your choosen coordinate frame...
– Sito
Apr 24 '19 at 10:51
• $\vec{r}$ is the position vector of the center of mass in space coordinate system.
– Eli
Apr 24 '19 at 11:24
• Yup. You have to use parallel axis theorem here Apr 24 '19 at 18:02