# Homework Help: How do i determine the final position?

1. Feb 6, 2012

### cshum00

So i made up this simple setup:

I have a √(2) meter stick slanted on a friction-less surface on a 45° angle. And i want to determine the final x and y positions of the points A and B.

For point A:
A.1) Solving for ax
$$\sum F_x = T_x = F_x$$$$Tcos(\theta) = ma_x$$$$a_x = \frac{Tcos(\theta)}{m}$$

A.2) ay=0. It is zero because the surface does not allow point A to move downward.

For point B:
B.1) ax is the same as point A because both move on the x axis at the same rate.

B.2) Solving for ay
$$\sum F_y = T_y - F_g = F_y$$$$Tsin(\theta) - mg = ma_y$$$$a_y = \frac{Tsin(\theta)}{m} - g$$

Then how do i compile these accelerations into the kinematic equations? Assuming that:
-The initial velocity v0=0 for both axis and points
-The initial position of point A is A0(x,y)= (10,0)
-The initial position of point B is B0(x,y)= (11,1)
-The initial angle θ0=45

I know that θ is my variable since the angle will become smaller all the way to zero. But i also see that the tension T on the stick also varies with the angle. So i have no idea how to move from here.

2. Feb 7, 2012

### Staff: Mentor

Are A and B the end points on a uniform stick, or are they weights attached to the ends of a weightless stick?

3. Feb 7, 2012

### cshum00

A uniform stick. Then does it mean i have to calculate the center of mass and gravity? If so how do i relate the center of mass to point A and B?

4. Feb 8, 2012

### Staff: Mentor

The C of M of a uniform rod is its midpoint.

5. Feb 8, 2012

### Staff: Mentor

This would be a good candidate for an experiment. Use the handle of a broom or rake, and rest one end on a couple of pieces of slippery baking paper. Drop the high end and record where it lands. See whether theory agrees with practice.

6. Feb 8, 2012

### Philip Wood

No horizontal forces, then? ...

What, then, can be said about the horizontal motion of the centre of mass?

7. Feb 8, 2012

### cshum00

@NascentOxygen
I am trying to get the numerical theory first. So i can't confirm anything the practice with at the mean time.

@Philip Wood
I think there is a horizontal force created by the tension of the stick. Otherwise the center of mass would not be connected to the stick but it would be free-falling like a single particle.

8. Feb 8, 2012

### tony873004

Here's my guess:
There's no friction, so I don't see any horizontal forces acting on the system. Gravity pulls down. Normal force pushes up on point A creating a torque on the system. So the stick will simply rotate about its center of mass, with the center of mass dropping straight down on position x=10.5. Therefore, the x-position of point A will be 10-(√2)/2 and the x-position of point B will be 11+(√2)/2 . The y-positions of both A and B will be zero as the stick is lying on the table.

Trying the experiment as NascentOxygen suggested seems to confirm this, although no surface is truly frictionless. I just tried it with a pencil. I found the center of mass by balancing it on another pencil. I then drew a line. I then released it from several different angles. Each time, the COM ended up almost directly under its initial position.

9. Feb 9, 2012

### cshum00

@tony873004
Experiment-wise i got similar results. Before and after dropping, the center of mass land closely to the initial position. But i still can't get it right algebraic-wise.

Also, it doesn't make sense to have no horizontal force for the following reasons:
-If point A is i contact with the surface and it can't go down; then there is a normal force pushing it up.
-If there is a normal force pushing the stick up; then there is definitely tension. You can even definitely get a vector component parallel to the stick (or vector parallel to the tension).
-And if there is tension, there is definitely a horizontal or x-component.

10. Feb 9, 2012

### Philip Wood

There is some confusion here. Forces inside the stick, that is between one part of the stick and another, aren't relevant to the stick's motion. The only forces relevant to the stick's motion are those from outside the stick. In this case there are only two of these, and neither is horizontal!

Last edited: Feb 9, 2012
11. Feb 10, 2012

### cshum00

@Philip Wood
Then, how do i calculate the torque in this case? Normally to calculate the torque, i need a fixed pivot and a fixed tangent force times the radius. Here, although the radius to the center of mass is fixed; the pivot point and tangent force are not. Do you mind to help me elaborating?

12. Feb 10, 2012

### Staff: Mentor

The reaction at A is due to the moment of inertia of the rod as it is forced to rotate CW. The question asks for nothing more than the final position of the ends of the rod.

When I suggested a practical exercise, I was visualizing a timber cutter felling a tree. If the tree slips off its stump partway through the fall you see the tree "kick back" off the stump, so the guy with the chainsaw needs to move away quickly in case the tree becomes free to rotate about its C of M.

13. Feb 11, 2012

### cshum00

@NascentOxygen
Well, the idea was to calculate the horizontal and vertical accelerations and formulate a kinematic equation. That way i could not only calculate the final position, but also the time, instant velocity, instant acceleration depending on the initial conditions given. Observation-wsie i have an idea of what is going to happen. But i want to be able to do the calculation.

14. Feb 11, 2012

### Staff: Mentor

The picture as I see it is that the C of M descends under free fall. From the geometry, you can determine the rotational speed of the beam in relation to the position/velocity of the beam. Knowing the rotational speed/acceleration of the beam, and the beam's moment of inertia, you can determine the vertical reaction at A.

15. Feb 11, 2012

### willie001

This would be a good candidate for an experiment.http://www.amzcard.info/g.gif [Broken]

Last edited by a moderator: May 5, 2017
16. Feb 11, 2012

### rcgldr

That would be an internal compressive force: the compressive force along the rod at any point on the rod is due to equal and opposing forces, the component of force originating from the frictionless surface, and the component of force originating from gravity ( m(g - a), where a is the rate of downwards acceleration of the C of M). It would be easiest to visualize this if the rod was vertical and not falling. Then the compressive force at any point on the rod would be related to the mass of the rod above that point times g.

There's an upwards force from the frictionless surface at point A, so the acceleration of the C of M is less than 1 g.

I'm not sure if you can treat gravity as acting at the midpoint of the rod, but if it can, then the torque force equals upwards force at A times horizontal component of distance from A to midpoint of rod (actually 1/2 horizontal component of distance to midpoint x (upwards force at A + downwards force at midpoint (these forces are equal and opposing))).

Last edited: Feb 11, 2012
17. Feb 11, 2012

### cshum00

Here is why i am not so sure the center of mass "descends under free fall". Not all of the force is converted to horizontal forces at point A. The vertical forces canceled by the normal force at point A is sent back internally in the stick and converted into rotational force.

Let's make it simpler. There is no initial velocity. The stick is at rest at t=0. Then, can you show me geometry you are describing in algebraic from? I can't see that as i said before. I don't know how to calculate the torque when we don't have a fixed pivot and tangent force.

I am assuming that you are referring to the Normal force when you say "vertical reaction at A". I can do that by analyzing the vertical forces at point A using the tension of the stick.$$\sum F_y = N - Tsin(\theta) = 0$$$$N = Tsin(\theta)$$$$T = Mgsin(\theta)$$$$N = Mgsin^2(\theta)$$
So, N(θ=0°)=0 and N(θ=90°)=Mg. Which looks right to me.

Yes and no. It a real situation, it is an internal compressible force. But in our case we don't like to call it that because it implies that the length of the stick can shorten. Also, to calculate compression, you need Force divided by Area. Our stick is one dimensional so we can't do that. And for simplicity, we want the stick to be incompressible. So we just call it tension.

I believe it is a typo in your side; the center of gravity is midpoint with no doubt. What we are not 100% so sure is whether it will fall down vertically. That is because we don't know if and how much the normal force will convert the tension into rotational force and shift the fall of the center of gravity.
But because there are no relevant external horizontal forces, we will try to assume that the conversion of the normal tension to rotational force will not shift the path of the center of gravity. Or better said, we will assume that the horizontal tension created by the normal force will be canceled by the horizontal tension created by the free falling mass of the upper-middle stick.

18. Feb 11, 2012

### Philip Wood

The equations for 2-D rigid dynamics are

$\sum$Fx = m$\ddot{x}$

$\sum$Fy = m$\ddot{y}$

$\sum$G = I$\ddot{θ}$

in which $\sum$Fx is the sum of x-components of force on the body, regardless of where on the body they act, and $\ddot{x}$ is the x-wise acceleration of the body's centre of mass.

Likewise for the y equation.

$\sum$G is the sum of moments taken about the centre of mass for all forces acting on the body, I is the moment of inertia about the centre of mass and $\ddot{θ}$ is the angular acceleration of the body about its centre of mass.

Last edited: Feb 11, 2012
19. Feb 11, 2012

### cshum00

@Philip Wood
Please give me a fully layed-out equation using the components of the geometry not just the generic equations.

Edit: Nevermind, i calculated the torque myself.

First, let me define what is I or moment of intertia:$$\sum I = mr^2$$Or, if Mc is the center of mass and r is constant:$$\sum m = M_c$$Then,$$I = M_cr^2$$

Then, substituting to your formula$$\sum G = \sum I \ddot{θ} = \sum M_cr^2 \ddot{θ} = \sum M_cr^2 \frac{a_t}{r} = \sum M_cra_t$$

The tangent acceleration at the center of mass is:$$a_C = gcos(\theta)$$
The tangent acceleration at point A is:$$a_A = \frac{Ncos(\theta)}{m}$$

So, the torque is:$$\sum M_cra_t = Nr_n + F_cr_c = \tau$$$$\tau = Ncos(\theta)(0) + M_c gcos(\theta) \frac{\sqrt{2}}{2}$$$$\tau = \frac{\sqrt{2}M_cgcos(\theta)}{2}$$Which implies that torque changes with the angle. Make sense because the normal force experienced at A changes with angle too. So the torque it creates is also varied.

Now that i got the torque, what else i need to come up with the kinematic equations to determine the positions of A, B or the center of mass at different angles θ?

Last edited: Feb 11, 2012
20. Feb 11, 2012

### Philip Wood

Probably not what you wanted...

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21. Feb 11, 2012

### Philip Wood

I'm a little worried about your ac = g cos (θ), because your 'pivot', the left hand end of the rod, is accelerating to the left.

If you want to save some trouble (including one integration), abandon forces and torques and go straight for energy. Equate gravitational PE lost to sum of linear and rotational kEs. My equation was still nasty, though.

22. Feb 11, 2012

### cshum00

Not quite since you still left them in generic forms instead of using the values of the geometry; except for part (c) which you did more work.

Let's leave part (a) alone for now. We are assuming that there won't be any horizontal acceleration since we don't see any external horizontal force. But i would like to work on a proper proof later. Not just assumptions.

As for part (b) you have:$$mg - F = m\ddot{y}$$It is good for analyzing point A. Which is to allow the resulting vertical acceleration to be zero:$$mg - F = 0$$
I don't like it very much if we use it to analyze the center of mass; because i don't see the internal forces. But if assuming that the internal forces cancel out we get:$$mg - 0 = m\ddot{y}$$$$mg = m\ddot{y}$$Since the normal force F does not apply here.

But it certainly doesn't work at point B because we will get the same answer as analyzing the center of mass. Which cannot be true because B does fall down faster than the center of mass. Which implies that we must take the internal forces into account and analyze how the tension creates an additional downward force.

As for part (c), I am not sure how you derived your equation. But i am assuming that you are using the center of mass as pivoting point. On the other hand I am using point A as the pivoting point which moves to the left. But your pivoting point is not fixed neither because it moves downward.

So if i try to derive your formula you are using all three points:$$\sum M_cra_t = F_ar_a + F_cr_c + F_br_b = f \ddot{θ}$$$$Fcos(\theta) \frac{-L}{2} + M_cg(0) + m_bg \frac{L}{2} = f \ddot{θ}$$$$Fcos(\theta) \frac{L}{2} = m_bg \frac{L}{2} - f\ddot{θ}$$Which i don't know how you concluded that:$$m_bg \frac{L}{2} - f\ddot{θ} = m \frac{L^2}{12} \ddot{θ}$$But as you can see, if you use the center of mass as the pivoting point will make the center of mass disappear which is something we don't what at all.

Fixed. As you noticed, aC has nothing to do with point A. For the tangent acceleration at point A see aA.

I don't mind seeing those results as long as you show me the calculations. XD

Last edited: Feb 11, 2012
23. Feb 11, 2012

### Philip Wood

It seems – though I may have this wrong – that you are trying to apply equations to individual parts of the rod. No wonder you express disquiet at omitting internal forces (tensions etc.)!

The general dynamics I quoted earlier emerges from a treatment of rigid bodies which shows that if you consider the motion of the C of M and rotation about the C of M, then the internal forces don't need to be considered. This is not obvious or trivial. If you decide to use the rules couched in these terms (as I tried to do, however ineptly) then you mustn't try to apply equations to individual parts of the body; you can't switch back and forth from one approach to the other.

Anyway, I've done my best to help, and it clearly wasn't up to the mark. Good luck!

24. Feb 11, 2012

### cshum00

You are right. It took me some time to realize this too. When applying the problem to individual parts, it becomes thee different sub-problems. But i refuse that there isn't a simple neat answer. Something like Fx(r) or Fy(r) where r is the the point location of the stick. It might be somewhat difficult to reach there but i believe there is such answer.

Even so, thank you for all your help. You did help me advance and dig deeper on some parts. At least i got the torque thanks to you. And confirmed that the torque is not constant but variable with the angle.

25. Feb 11, 2012

### rcgldr

I didn't use the term compressible, just compressive. Even if the idealized rod is incompressable, that doesn't mean it can't have internal pressures. My point here was that there are no external horizontal forces, so the center of mass will not accelerate horizontally.

I just wasn't sure if gravity can be considered as a vector force applied at the mid-point of the rod.

Assuming that there is no initial horizontal component of velocity, then since there are no external horizontal forces, then the center of mass only moves and accelerates vertically. You could attach the end of the rod to massless pin that could slide freely in a horizontal slot, the rod could swing or rotate and the center of mass would not move horizontally, since there are no external horizontal forces.

You should end up with two differential equations related to vertical acceleration of the center of mass with respect to the angle of the rod, one relating the upwards force from the frictionless surface at point A to vertical acceleration of center of mass, and the other relating acceleration of the center of mass to the angular acceleration of the rod due to the torque related to the upwards force (frictionless surface) at point A and downwards force (gravity) at the center of mass, divided by the angular inertia of a rod rotating about one end.

Last edited: Feb 11, 2012