# Slipping stick

1. Jan 23, 2005

### ceptimus

A long thin straight stick is balanced on its end, on a flat horizontal frictionless surface (like an ice rink).

The balance is not quite perfect, so the stick slips / falls over. Clearly, its centre of mass will go in a vertical path down, to hit the surface at the point the stick originally stood on.

Question: Will the bottom of the stick leave the surface before the top of the stick touches it?

Last edited: Jan 24, 2005
2. Jan 24, 2005

### Gokul43201

Staff Emeritus
I have a feeling that the torque on the rod may very well require the bottom end to jump off.

3. Jan 24, 2005

### ceptimus

If it does leave the surface, what is the angle between the stick and the surface at the instant it leaves?

4. Jan 24, 2005

### vikasj007

we can assume that the rod is uniform, and the centre of mass is at the centre of the rod. so if the centre of mass goes vertically down to touch the point where the base of the stick was, then the centre of the rod has to move vertically down, so the bottom end should leave the surface as soon as the top end starts to move downwards, just as if the centre is fixed and the rod is rotated, but just that here with that, the rod is simultaneously moving downward too.

you'll get it if you analyse the motion of the rod in 2 different parts.
first of the rod moving down, and second of the rod rotating.

5. Jan 24, 2005

### Bartholomew

Does it make sense to speak of a torque when there is no fixed center?

If the rod were in outer space and you gave it a tap perpendicular to one end of it, would it spin in place, or move, or some combination of the two? If it spins in place how is this consistent with conservation of momentum?

6. Jan 24, 2005

### gerben

I think the bottom of the stick will not leave the surface before the top of the stick touches it.

If the bottom would leave the surface there would be no reason for the stick to rotate (no torque), the stick would just fall straight down and not rotate. Already when the bottom of the stick and the surface would have separated an infinitesimal distance the stick would fall straight down back to the surface, so it can never leave the surface.

7. Jan 24, 2005

### Kittel Knight

Good question!
My feeling was it would jump off.
The condition to be satisfied is : the angular acceleration (derived from the energy equations) should be negative at some point.
But, after some calculations, I've concluded it wont leave the surface.

8. Jan 25, 2005

### K.J.Healey

Assuming that theres no friction, and theres no distortion of the stick, then the whole stick would hit the ground at the same time, right? Any angular acceleration would have to be about the point in contact with the surface, the only semi-stationary point, there is no torque possible anywhere else.

9. Jan 25, 2005

### ceptimus

That's right, but there was the possibility that the rotation induced in the stick would carry the lowest point of the stick up, faster than gravity carries it down. In fact I thought that this probably would be the case, until I did the math, and like Kittel Knight found that it isn't.

That's why I posted it here. I thought it was a simply stated problem but with a (for me anyway) non-obvious answer that required quite a bit of thought and calculation to fully analyse.

Last edited: Jan 25, 2005
10. Jan 25, 2005

### ceptimus

Good question. This bothered me too.

The answer is that some angular momentum is imparted into the surface that the stick is standing on. It helps to visualise the stick standing on the centre of a (fairly) light horizontal tile. As the stick falls, the contact point on the tile moves away from the tile's centre. The off centre force on the tile causes it to gain angular momentum equal and opposite to that the stick gains.

11. Jan 25, 2005

### Bartholomew

Well, okay, but that wasn't really what I was asking about. If the stick is in outer space and you throw a ball at one end of it to strike at a right angle, how is momentum conserved? The ball's loss of momentum must be balanced by the stick moving forward at an appropriate speed, and how does this leave anything left over for rotation of the stick, which intuitively should happen? But never mind, I think I figured that out.

12. Jan 25, 2005

### K.J.Healey

The impact would create both a rotation and a translation. I could dig up my dynamics book somewhere for the proof, but it does.

13. Jan 25, 2005

### Bartholomew

Yes, the momentum would be conserved but not the energy, and the leftover energy would go into rotation.

14. Jan 26, 2005

### Kittel Knight

In order to clarify, I meant some point along the time.

Besides that, the sticks center follows a vertical path, and the induced rotation is around the center.

15. Jan 26, 2005

### Bartholomew

Well, there is a possible reason for the stick to leave the ground. At all times before it leaves the ground, it must rotate fast enough to keep up with its downward motion. It will never reduce its rotation speed. But as time goes on, a given rotation speed becomes increasingly efficient at moving the tip of the rod away from the ground--if the rod is rotating at 90 degrees per second and _not_ falling, its tip is moving away from the ground faster when the angle between rod and ground is 30 degrees than when the angle is 60 degrees.

If the rod fell at a constant rate, this effect would make the tip of the rod leave the ground immediately. But since it accelerates as it falls it is more complicated.

16. Jan 26, 2005

### Bartholomew

If you could figure out how the rotation and falling speed would be related when the end of the stick is "locked" to the ground, then you just look for the place where rotation speed has to decrease for the stick to remain on the ground. At that point, the stick leaves the ground. Dunno how to do that though.

17. Jan 26, 2005

### Bartholomew

Okay, I think it never leaves the ground.

mg(L / 2 -h) = 1/2 (1/12) m * L^2 * w^2 + (1/2)m*v^2
m is the mass, g is the force of gravity, h is the height at present of the center of the stick above the ground. (L / 2 - h) is how far the stick has fallen. L is the length of the stick, w is the angular speed (in radians) the stick is rotating at, and v is the speed of the stick's center towards the ground. 1/12 m * L^2 is the moment of inertia of the rod about its center.

This is because the gravitational potential energy must be converted as the stick falls into a combination of rotational and linear kinetic energy.

v (which is downwards) must equal the upwards component of the rotation at the stick's lower end so that that end can stay at ground level. By drawing a diagram, it can be seen that v = w * cos a, where a is the angle the stick makes with the ground. Here I assume that L = 2 so that the "radius" of the stick can be 1 so that w can just be the linear speed at the tip. Some more drawing demonstrates that h = sin a, or a = arcsin h (if L is 2). v = w * cos arcsin h, and cos arcsin h = (1 - h)^.5, so v = w * (1 - h^2)^.5

Plugging this into the previous equation (assuming L is 2) it works out to
w = (g(1 - h)/(2/3 - h^2 / 2))^.5

Graphing this on a calculator from 0 to 1 shows that it is decreasing. Since h is actually backwards with respect to time, w increases continually through the time span, so the stick never leaves the ground.

Last edited: Jan 26, 2005
18. Feb 2, 2005

### hemmul

It's clear, that thee are only 2 forces in the problem: gravitational force mg, applied to the whole stick (but in calculations we simplify this by applying it to the mass centre) and the reaction force N that is normal to the surface (no friction). Now, the fact that the stick leaves the ground means that N=0.
well, this post consists of a brain teaser and a question :)
brain teaser: find out whether N is becoming zero during the fall.
question: i received that N=mg/4 - during the entire process! So it's constant... isn't that weird?

19. Feb 2, 2005

### ceptimus

I don't think that's right hemmul.

Consider the start position where the stick is balanced on end. Let's pretend the balance is so good that the stick stays like that for five seconds before it begins to visibly slip / fall. Of course, once it starts to fall it all happens fairly quickly - though smaller sticks will fall over faster than larger ones.

Now during that inital five seconds, the reaction force (N as you call it) must be almost exactly equal to mg, from then on it reduces (else the stick would not fall), and if, as you rightly say, it ever goes negative, then the stick would leave the surface. It seems that most of us that have done the math now agree that it never does go negative though.

20. Feb 3, 2005

### hemmul

yes, you're right, thanks. i realized, that the error in my calcs was in that i considered the rotation point stable. but actually it is moving nonuniformly along the surface... now i realized that i'm beginning to forget the rotational body theory, so please don't hesitate to correct me if something is wrong:
the variables in torque equation (angular momentum = moment of inertia * angular acceleration) are calculated relatively to the rotation point/axis that is at rest. but what if it is moving? say, at a constant acceleration? how are we supposed to write down the rotation equation?