# Tricky problem: Falling plank originally leaning against a wall

1. Aug 1, 2011

### Order

1. The problem statement, all variables and given/known data

A plank of lengt 2l leans against a wall. It starts to slip downward without friction. Show that the top of the plank loses contact with the wall when it is at two-thirds of its initial height.
Hint: Only a single variable is needed to describe the system. Note the motion of the center of mass.

2. Relevant equations

$$\tau=I\alpha$$
Energy conservation

3. The attempt at a solution

I dont get very far at all.

First of all I evaluate the torque of the system: $$\tau=lMg\sin \theta=I\ddot{\theta}=\frac{Ml^{2}}{3}\ddot{\theta},$$ which leads to $$\ddot{\theta}=\frac{3g}{l}\sin \theta.$$ This is not possible to solve exactly and therefore is of no help.

Next, I try energy conservation to evaluate the angular velocity at a height of two-thirds of the original: $$Mgl\cos\theta=\frac{1}{2}\frac{Ml^{2}}{3}\omega^{2}+\frac{2}{3}Mgl\cos\theta,$$ which leads to the result $$\omega^{2}=\frac{2g}{l}\cos\theta.$$ Now this is of no use either since it cannot be combined with the first equation. It is a relation relative to the original angle, whereas the first equation above is a general differential equation.

Now if neither torque evalutations nor energy conservation leads anywhere, I dont know how to go on. Does anyone have a hint?

2. Aug 1, 2011

### SammyS

Staff Emeritus
Draw a free body diagram.

You are finding the torque about which point?

3. Aug 1, 2011

### Staff: Mentor

It's a tricky problem. Use the hint: What path does the CM follow as the plank falls?

4. Aug 2, 2011

### Order

Here is a FBD as I think it should be. In the diagram I also think that the normal forces are $$N_{1}=Mg\sin\theta$$ and $$N_{2}=Mg\cos\theta$$ where the latter might be most interesting. See my post below for further thoughts.

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5. Aug 2, 2011

### Staff: Mentor

How did you determine this?

6. Aug 2, 2011

### Order

Sure, I can try that. First of all I can evaluate the center of mass in the x direction. $$x_{cm}=l\cos\theta$$ and differentiating this gives $$\frac{dx_{cm}}{dt}=-l\sin\theta\omega$$ Now since theta is evaluated in the positive direction, whereas I am interested in the negative direction, I revise this to $$\frac{dx_{cm}}{dt}=l\sin\theta\omega$$ Differntiating once again yields $$\frac{dx^{2}_{cm}}{dt^{2}}=l\cos\theta\omega^{2}+l\sin\theta\alpha$$ Now putting in relations for alpha and omega in my original post, slightly changed, I get $$\frac{dx^{2}_{cm}}{dt^{2}}=6g\cos\theta(\cos\theta-cos\phi)-3g\sin^{2}\theta,$$ where phi is the starting angle.

At first I thought Í could find where the acceleration is zero to find where the plank loses contact, but I find that the acceleration is always positive. Instead I might use the results from my free body diagram in post above to solve the equation $$M\frac{dx^{2}_{cm}}{dt^{2}}=N_{1}=Mg\cos\theta$$ which leads to the equation $$6\cos\theta(\cos\theta-\cos\phi)-3\sin^{2}\theta=\cos\theta$$ to get a relation between theta and phi. Unfortunately it seems like a very difficult equation to solve.

I can do the same thing with the center of mass in the y direction. Now the result is $$y_{cm}=l\sin\theta$$ $$\frac{dy_{cm}}{dt}=l\cos\theta\omega$$ $$\frac{dy^{2}_{cm}}{dt^{2}}=l\sin\theta\omega^{2}-l\cos\theta\alpha$$ But unless my equation I found above is correct, I am stuck once again.

7. Aug 2, 2011

### Staff: Mentor

Describe in words the path of the CM. It's surprisingly simple. (The path, not the problem.)

8. Aug 2, 2011

### Order

On closer thought I need to revise this to $$N_{2}=\frac{\sin\theta\cos\theta}{2}$$ where the trignonometric part is due to the force being applied perpendicular to the gravity force through the plank. The factor 2 is due to the normal force being at twice the length. I think this should be correct.

9. Aug 2, 2011

### Order

I don't know how this helps. Will think about it. But as for know I can readily see that it is a circle.

10. Aug 2, 2011

### Staff: Mentor

Excellent! Now apply conservation of energy to find the velocity of the CM as a function of angle.

11. Aug 2, 2011

### Order

I still have no clue why this is good. I try to think it through and although it seems like something good I cant see the end of the string.

I can express it as a function of a fraction of the original height, or, as I guess you wanted me too, express it as a difference of the original and the final angle: $$v^{2}_{CM}=2lg(\sin\phi-\sin\theta)$$ or if a is a fraction of the original height $$v^{2}_{CM}=2lg\sin\phi(1-a)$$

12. Aug 2, 2011

### Staff: Mentor

There's only one angle here: The angle the plank makes with the wall. (Assume it starts out perfectly flat against the wall, and is then given a nudge.)

To apply conservation of energy, you must include both rotational and translational KE terms.

13. Aug 2, 2011

### Order

Ooops, sorry about that. In that case i san write $$v_{cm}^{2}=2gl(1-\sin\theta)-\frac{1}{3}l^{2}\omega^{2}$$ or $$(\omega l)^{2}=\frac{3}{2}gl(1-\sin\theta)$$

By the way, i'm off tomorrow, so maybe I need to wait a week to solve this problem if I don't today.

14. Aug 2, 2011

### Staff: Mentor

Good. Hint: Consider the horizontal component of this velocity.

15. Aug 9, 2011

### Order

Ok, now I'm back from my little vacation. I still don't know how velocities can help, though. I want to deal with accelerations. So I need to differentiate the equation above.

Firstly I separate the horizontal and vertical velocities.

$$\omega^{2}(\sin^{2}\theta+\cos^{2}\theta)=\frac{3}{2}gl(1-\cos\theta)$$ I can then differntiate this to $$-2\omega\sin\theta a_{x_{CM}}l+2\omega\cos\theta a_{y_{CM}}l=-\frac{3}{2}gl\cos\theta\omega$$ Continuing this way, first dividing by omega, then substituting for axcm=N2 (where it loses contact), omega2 (angular velocity) and alpha (angular acceleration), I get the equation $$\sin^{2}\theta+2\sin\theta\cos\theta-\frac{3}{2}=0$$ This leads to the result that the plank loses contact at about 5% of its initial height (analyzing the equation numerically).

Am I on the right track or am I supposed to do something completely different?

16. Aug 9, 2011

### Staff: Mentor

Yes, you'll need to differentiate.

Start by giving an expression for the horizontal component of the velocity. Then differentiate that.

17. Aug 9, 2011

### Order

Yes, I needed the free body diagram to reevaluate the angular acceleration, and using that the horizontal acceleration should be zero I arrived at the equation $$-2\sin\theta+2\sin^{2}\theta-\cos^{2}\theta=-1$$ with the non-trivial solution $$\sin\theta=\frac{2}{3}$$ Thanks for help Doc!

18. Aug 9, 2011

### Staff: Mentor

Excellent! (A tricky problem indeed.)