# Flight of a pencil

1. Mar 15, 2008

### Lookaash

Hello everyone, I am dealing with a tough problem. I place a pencil on the edge of a table and I hit it. I want it to fly as far as it is posible. I dont really know how to describe its path - how much will the the hitting cause its rotation and how much will it give the pencil velocity?
There are several hidden factors that are making the problem difficult. One is that when I hit the pencil, I continue on my path, so it is just not an impulse. Another is the width of the table - the pencil jumps of it.

Basically, the only thing I know for sure is that if I place the pencil on a table that has no width so that the center of mass is directly above edge of the table, then if I hit it very near of the center of gravity, it will just rotate and will not fly at all.

I know that the length of the flight depends on how far is the center of mass from the edge of the table and also on where and how I hit it. I am getting really confused, though. Please, is there anyone who could help me?

2. Mar 16, 2008

### Lookaash

Any suggestions?

3. Mar 16, 2008

### John Creighto

The pencil should have linear and rotational velocity when it leaves the table. The pencil will rotate about the point where it touches the edge of the table. The amount of angular velocity you give it should be roughly related to how hard you strike it. Knowing the point the pencil rotates and the angular velocity you should be able to calculate the linear velocity.

4. Mar 16, 2008

### Lookaash

I am not getting how to calculate the linear velocity..

5. Mar 16, 2008

### John Creighto

Well, the linear velocity is how fast the center of mass moves. If the pencil is not spinning abut the center of the pencil then the center of mass will be moving.

Using basic trig the center of mass will move with a velocity of:

$$v_{cm}=d \omega$$

Where:
d is the distance from the center of the pencil to the edge of the table (in metters)
$$\omega$$ is the angular velocity of the pencil in radians per second.
$$v_{cm}$$ is the linear velocity of the pencil in meters per second.

Last edited: Mar 16, 2008
6. Mar 16, 2008

### wysard

I always sorta wondered about that. I thought the distance from the edge of the table would have an impact (no pun intended) on the azimuth angle the pencil leaves the table at. Given the OP was looking for distance then linear velocity is only half the puzzle, the other being angle. Given that a pencil leaving at high speed at 90 degrees to the table goes up and then down with little distance and one that leaves at good speed at 0 degrees still does not go as far as one leaving at 45 degrees at moderate speed. When we add that both the pencil and the table top are both elastic solids I would be surprised if we can get much better that a polynomial equation that describes a best guess scenario. I look forward to reading more.

7. Mar 16, 2008

### John Creighto

We can always be more precise if we want but from my observations the pencil takes off at a pretty close to horizontal angle.

8. Mar 16, 2008

### wysard

True generally. I find, from a couple of ad hoc-experiments just now that the azimuth seems to depend to some degree on what percentage of the pencil is hanging over the edge and how far from the end of the hanging over portion I hit the pencil. Could a ratio of the mass of pencil on the table to the mass between the edge and the striking point play a role in how the force is transferred? I would expect that if say half the pencil hung over and you hit it on the end the pencil would be just about horizontal, but if you put say 80 percent on the table and hit it on the end because the center of mass of the pencil is well behind the fulcrum of the edge of the table a larger percentage of the force is transferred into vertical movement. It's outside my area but I am curious.

9. Mar 16, 2008

### John Creighto

Initially when you strike it if only a small part is over the edge then there will be more vertical velocity but, as the pencil rotates it wraps around the edge and some of that rotational velocity is transfered into horizontal velocity.

10. Mar 16, 2008

### wysard

I got it. I have fat hands. I just tried it with the back edge of a butter knife. The difference is in the carry through. If, whether with my hand or the knife edge I carry through the strike it is pretty much horizontal. If it is a sharp sudden rap it is not. I can only surmise that when I rap the pencil the only forces in play are the masses involved and the ratio of thier distance from the fulcrum and centre of mass. If I follow through the result is horizontal both ways from which I realize that whatever horizontal motion might have happened, by continuing to force the pencil down I actually slide down towards the point (Assuming the eraser is on the table and the point over hangs the edge) and because of the constant sliding force I constrain the angle of moment to be forced to the edge of the table regardless of the percentage of the pencil on the table.

11. Mar 17, 2008

### tiny-tim

… reaction force must be zero …

It is tempting to think that the pencil moves away from the table because of the reaction force from the table (rather like something bouncing off a wall).

But I think it's the opposite - the pencil leaves the table when the reaction force is zero, because that's more-or-less the definition of losing contact.

We know that the centre of the pencil will follow a circle whose centre is the edge of the table.

So the question is, at what combinations of angle of the pencil, angular velocity, bendability of the pencil, and point of application and direction and strength/mass of the force, will the reaction force be zero?

12. Mar 17, 2008

### wysard

It will? I thought that in free flight the pencil would orbit, or circle, around it's centre of mass. If the pencil is half on and half off the table, they would be the same, but if what you imply were categorically true there would be no need for levers.

13. Mar 18, 2008

### tiny-tim

Hi wysard!

I was only talking about the situation before the pencil leaves the table: if there is no slipping, then the whole pencil rotates about the edge of the table until the reaction force is zero.

From that moment, the pencil loses contact with the table, and, as you say, it continues to rotate indefinitely round its centre at the same angular velocity which it had at that moment around the edge of the table (and of course its centre follows a parabola starting with the velocity which it had at that moment).

14. Mar 18, 2008

### wysard

Hey tiny-tim!

That is exactly the point I was making in my earlier post about contact time. If you sharply rap the pencil the energy is transferred quickly and because the contact time with the knife is very short the pencil leaves the table before the pencil rotates very much around the table edge leading to a high arc (assuming again that most of the pencil is on the table). If you carry through the pencil is constrained to rotate in a much larger arc around the edge of the table before it comes free resulting in a much lower flatter arc.

Excellent description of how to find out exactly what arc the pencil will travel in! If you transfer the same energy to the pencil, on way quickly with little contact, and the other slowly with longer contact the pencil must perforce describe two distinctly different parabolas no? Or did I miss something?

15. Mar 18, 2008

### tiny-tim

… moment of inertia … help … pleeeze … !

Hi wysard!

I think it's about time we started doing some calculating. :yuck:

Suppose the pencil has mass M and length 2a, and its centre C is distance r from the edge E of the table, at which the reaction force perpendicular to the pencil is R, and that E is distance b from the point of application P of a variable force F.

Suppose that the pencil is constrained to pivot about E, and to move so that at time t its angle to the table (which is is on the right) is µ(t), anti-clockwise.

Then the applied forces perpendicular to the pencil are Mcosµ F and -R; so, applying good ol' Newton's second law in that direction, Mcosµ + F - R must equal the rate of change of momentum in that direction, which is M times the rate of change of the velocity of C, which is -r.d^2µ/dt^2.

In particular, when the pencil loses contact, R = 0, and so:
Mcosµ + F = -M.r.d^2µ/dt^2.​

It is remarkable that, for contact to be lost, there must be a large angular deceleration! Essentially, this is because P must be forced to move slower than C is "expecting"!

Now we need to find F, which we can do by taking moments about either C or E …

erm … I never really got the hang of moments of inertia …

… would someone else help by finishing this off for me … please!!

16. Mar 19, 2008

### wysard

I love math! It's like playing with a God's LEGO set...

This is, to me, the salient bit. No one, at least here, will argue something silly, there is the crackpot list for that...

So, what causes P to move slower, thereby causing an angular deceleration? How about, you stop pushing the pencil down? Then Mcosµ + F = -M.r.d^2µ/dt^2 and the pencil, due to lack of a counter force P is no longer constrained to rotate about E. So if P is a momentary force and not a continuing force ie: a rap where P is applied for, say, 30 degrees of rotation, versus an ongoing push where P continues through "t" until the pencil is 90 degrees to E at which point R reaches zero for a perpendicular force giving the expected horizontal flight?