Solving a Pole-Leaning Problem with Negligible Mass

[Born]

Homework Statement

A pole of negligible mass leans against a wall, at angle θ with the horizontal. Gravity is directed down.

(a) Find the constraint relating the vertical acceleration of one end to the horizontal acceleration of the other.

(b) Now suppose that each end carries a pivoted mass M. Find the initial vertical and horizontal components of acceleration as the pole just begins to slide on the frictionless wall and floor. Assume that at the beginning of the motion the forces exerted by the rod are along the line of the rod. (As the motion progresses, the system rotates and the rod exerts sidewise forces.)

Homework Equations

$$(b-y)^2+x^2=L^2$$

The Attempt at a Solution

I feel that the constraint is given be taking the second derivative of the previous equation and getting back $$ \ddot{x}x + \dot{x}^2 = \ddot{y}y + \dot{y}^2$$

The problem comes when analyzing the forces. I get

$$\hat{j}: Mg-F_{ry}=M\ddot{y}$$
$$\hat{i}: F_w-F_{rx}=0$$

For the mass falling vertically and

$$\hat{j}: Mg+F_{ry}-N=0$$
$$\hat{i}: F_{rx}=M\ddot{x}$$

for the mass on sliding on the floor (where F_r is the force of the rod; N, the normal force; and F_w, the force of the wall).

All help is welcome. Thanks in advanced!

 

[Gil]

The time derivative of $x^2$ is $2x\dot x$. The second derivative is $2 (\dot x^2 + x \ddot x)$.

 

[Born]

Yes and the same for y so I cancel the 2.

 

[Gil]

I feel that the constraint is given be taking the second derivative of the previous equation and getting back $$ \ddot{x}x + \dot{x} = \ddot{y}y + \dot{y}$$

The $\dot x$ in your expression is linear when it should be squared, that's what I mean.

 

[Born]

Oops. Sorry, didn't notice that slipped out of the equation. In any case, the real problem here is the analysis of the forces. Any ideas?

 

[DocZaius]

I would just concentrate on solving for $\ddot y$ and let the constraint lead you to $\ddot x$. What do you know to be the initial velocities of $x$ and $y$? Since you are solving only for initial acceleration, you should put that in your constraint as that would simplify things. Then once you solve for $\ddot y$, you've got $\ddot x$ for free. Now consider the two forces acting on the top mass. You've got gravity, but you've also got this force directed up and to the left transmitted from the ground. Surely it depends on the angle?

 

[Born]

Ok so if I solve for \ddot{y} using the equations I have already I get: 2g-\frac{N}{M}=\ddot{y} but it's not very useful since I don't know N and the only way to figure it out is finding F_r which I can't figure out how to get. I am aware I have an angle with respect to the horizontal and the weight as part of the tringle but the analysis isn't giving anything satisfactory.

 

[Born]

I really wish to just use $Mg$ as the force that is pushing down on the pole such that $F_r= Mgcos(\theta)$. I guess this is true. It just feel awkward since this would imply $M\ddot{x}=F_rcos(\theta)=Mgcos^2(\theta)$. The only thing I'd have to find then would be how $\theta$ changes in time.

 

[haruspex]

I really wish to just use $Mg$ as the force that is pushing down on the pole such that $F_r= Mgcos(\theta)$.

Theta is the angle to the horizontal, and F_r is the compressive force along the rod, right? First, what angle does the gravitational force make to the rod? Second, your equation has no acceleration term.

 

[Born]

Oops. Well, I see that the angle of gravity to the rod is $\frac{\pi}{2}-\theta$. I think a picture would better explain how I see it. This would mean the acceleration in $y$ can be found by $$Mg(1-cos(\frac{\pi}{2}-\theta)sin(\theta))=Mg(1-sin(\theta)sin(\theta))=Mg(1-sin^2(\theta))=Mgcos^2(\theta)=M\ddot{y}$$

 

[haruspex]

the acceleration in $y$ can be found by $$Mg(1-cos(\frac{\pi}{2}-\theta)sin(\theta))$$

You seem to be assuming Fr = Mg cos(θ). How do you get that?

$$Mgcos^2(\theta)=M\ddot{y}$$

Sanity check: does that give the right answer when the rod is vertical?

 

[Born]

I - I got $F_r=Mgcos(\frac{\pi}{2}-\theta)$ from the diagram in the picture in the last post.

II - I'm very sorry if this sounds dumb, but it does. When $\theta=\frac{\pi}{2}$, $\ddot{y}=0$. Also, the acceleration is greatest when the bar is horizontal and $\theta=0$ making $cos^2(\theta)=1$ and therefore $\ddot{y}=g$. Why is this wrong?

 

[haruspex]

I - I got $F_r=Mgcos(\frac{\pi}{2}-\theta)$ from the diagram in the picture in the last post.

Yes, sorry, I meant you are taking Fr to be Mg sin(θ). (I would tend to work in terms of angle to the vertical, so I get a bit confused in this thread sometimes.)
Now, I'm not saying it's wrong (in fact I believe it is right), but I don't understand how you deduced it. Mg is a force acting on the mass M. It does not directly act on the rod. Since the mass is accelerating, it is not obvious what force is exerted on the rod by the mass. Were it held static at the bottom end of the rod, you would have Fr = Mg cosec(θ).

II - I'm very sorry if this sounds dumb, but it does. When $\theta=\frac{\pi}{2}$, $\ddot{y}=0$. Also, the acceleration is greatest when the bar is horizontal and $\theta=0$ making $cos^2(\theta)=1$ and therefore $\ddot{y}=g$. Why is this wrong?

Me getting theta the wrong way again . You do seem to have found the right answer, I just don't understand how.

 

[Born]

Sorry I've taken so long to respond.

To be honest with you, I just "kepplered" it. It was a hit-and-miss problem-solving. I also found it weird at first that the mass would apply it's force in such an odd way, but after thinking about the system as a pivoted mass that was in fact connected to the pole and restrained to a certain motion and doing some (much simpler) demos of this set up I was quite pleased with the previous answer. Another good sign is the equation for the acceleration of the second mass, Which I derived as follows

$g\cos(\theta)=\ddot{y}$ and in this scenario since $\dot{y}=0$, $\ddot{y}y=\ddot{x}x$ $\therefore$ $\ddot{y}=\ddot{x}\cot(\theta)$

Substituting $\ddot{y}$ I get: $g\cos(\theta)=\ddot{x}\cot(\theta)=\ddot{x}\frac{\cos(\theta)}{ \sin(\theta)}$

And finally solving for $\ddot{x}$,

$\ddot{x}=g\sin(\theta)$

 

[Born]

Wow total brain fart. $\ddot{y}=g\cos^2(\theta)$ NOT: $\ddot{y}= g \cos(\theta)$

Therefore $\ddot{x}=g\sin(\theta)\cos(\theta)=\frac{g\sin(2\theta)}{2}$

So, I'm back were I started. The way the equation behaves is pretty interesting though, with a max value of $\frac{1}{2}$, it starts at zero and returns when $\theta=\frac{\pi}{2}$