- anonymous

Consider a straight stick standing on end on (frictionless) ice. What would be the path of its center of mass if it falls?
I think that it will be a vertical line, but can't convince myself. Can anybody explain me this phenomenon?

- schrodinger

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- anonymous

@Vincent-Lyon.Fr

- anonymous

- UnkleRhaukus

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## More answers

- UnkleRhaukus

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- UnkleRhaukus

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- UnkleRhaukus

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- UnkleRhaukus

then bounce

- anonymous

Yes I know that. But why it falls like this?

- anonymous

@ArchiePhysics , totally depends on where you apply the external force to displace the rod

- UnkleRhaukus

well assuming the stick was just tipped over, not pushed , the center of mass of the stick will obey conservation of momentum in forward-backward direction,

- anonymous

@shivam_bhalla There is no any external force except the force of gravity

- anonymous

@ArchiePhysics , Why would it fall without external force ??

- anonymous

It doesn't matter. May b some tiny-winy fluctuation occurs in the field of gravity, which you can neglect.

- anonymous

@ArchiePhysics , It matters. We are assuming ideal conditions to solve problems. So without applying external force, the rod can't fall. 2nd, the rod won't fall if the external force is acting perpendicular to the rod on its centre of mass

- anonymous

@UnkleRhaukus Ok if it so, why the center of rotation of stick is the center of mass?

- anonymous

@shivam_bhalla OK, assume that there was initial velocity perpendicular to the stick the force was applied to the top what does it change?

- experimentX

circle??

- anonymous

@experimentX I don't think so. Because there is no friction which will hold the stick at the tip.

- experimentX

Oh ... i didn't see that

- UnkleRhaukus

assuming the stick is of uniform density , the center of mass will be in the center of the symmetrical stick

- anonymous

@ArchiePhysics , the centre of mass will not shift along the horizontal direction and only shift down vertically. So @UnkleRhaukus is right in this case

- anonymous

@UnkleRhaukus I agree, what next? Why it should the center of rotation?

- anonymous

Force is applied to the center of stick and the arm of the force is half-way between the tip and the center.

- anonymous

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- anonymous

@ArchiePhysics , what makes you say that. On the contrary, the rod will not slip and continue to move along without falling

- anonymous

@shivam_bhalla Because it rotates, so there must be some point of rotation

- anonymous

@Avva , one question to you. Will your door turn if you apply force at the hinge of the door perpendicularly ??

- anonymous

- anonymous

Of course won't.

- anonymous

But the stick is rotating, isn't it?

- anonymous

How do you know the point of rotation of a stick @ArchiePhysics ?

- anonymous

I don't know where is it, I'm just guessing.

- experimentX

I think we need to find the axis of rotation first!!

- anonymous

@experimentX , exactly !

- anonymous

+1

- experimentX

Consider a straight stick in space ... if we apply a force ... at any point on stick perpendicular to it's length ... then what would be it's axis of rotation??

- anonymous

It will begin to rotate and may be to move forward.

- anonymous

Torque T about cm of rod
\[t= r * F = i _{cm} (\alpha)\]

- anonymous

where r= L/2 i
i_cm = Ml^2/12
alpha--> angular rotaion

- anonymous

Why about center of mass? It should be about center of rotation.

- anonymous

Second, you need to apply conservation of angular momentum to find point of rotation

- anonymous

- anonymous

@ArchiePhysics , I specified the formula specifically for torque about CM

- anonymous

I don't need formulas I know them I want to know the only thing where is the center of rotation and why is it there.

- anonymous

Sorry I recognised that later.

- experimentX

lol ... try rotating things here
http://mrdoob.com/projects/chromeexperiments/google_gravity/

- anonymous

@experimentX Great stuff, but anyway, let's go on)))

- anonymous

Sorry for the statement "Second, you need to apply conservation of angular momentum to find point of rotation" because if no net external torque acts on a system, the total angular momentum of the system remains constant.

- anonymous

It's ok, although external torque is there, the force of gravity is applied to the center of mass, isn't it?

- anonymous

Found the solution here to your problem @ArchiePhysics
-->http://www.physicsforums.com/showthread.php?t=281440

- anonymous

Great let's see that!

- anonymous

if a straight stick falls in a frictionless ice, does it make a sound?
if the still simply topples over, there is some angular momentum created, some external force was applied, where was it applied affects the answer. If you assume the stick was given a torque about the center of mass, which becomes the center of rotation, then your answer is correct. State this assumption and answer as you thought.

- Vincent-Lyon.Fr

Ok, this problem was discussed earlier here. The actual question was: find the point of the rod whose path is circular.
It is a simple geometry problem once you get rid of some easy dynamics.
Once pushed out of (unstable) equilibrium by infinitesimal action, the only two external forces acting on the rod are its weight and the normal force of the plane.
Both forces are vertical, hence by N's second law, no horizontal acceleration of centre of mass.
As it starts its motion form rest, the centre of mass will move vertically, that's all.
Now, the second part (finding P with a circular path) is trickier although pure geometry. I have this on rough paper by two methods, but I like to post clean things onto OS, with clear handwriting ;-) , which needs some more time.

- Vincent-Lyon.Fr

Link to similar problem:
http://openstudy.com/users/vincent-lyon.fr#/updates/4facbd35e4b059b524f86008

- anonymous

The thing that I'll show you may seem to you ridiculous but that was what I needed.
I wanted somebody to prove me, or explain me that the center of mass would be the center of rotation in the given case.
@Vincent-Lyon.Fr you are absolutely right saying that
"Both forces are vertical, hence by N's second law, no horizontal acceleration of centre of mass."
But that was not that obvious to me until I convinced myself using geometry).
|dw:1337945093637:dw|
I got rid of the force of reaction and the remnant forces are shown.
The force F2 that pulls the center of mass to the left is equal in magnitude and opposite to the horizontal component of F1, so there is no any motion there, relative to the horizontal axis, so it gives us the axis of rotation not the point because the center of mass moves relative to the vertical axis of coordinates. To calculate the rotational motion alone we must use P as a central point of rotation because if there were strong enough force of friction the stick would have rotated about it.
Thanks everybody, for the help you gave me, I really appreciate it!

- Vincent-Lyon.Fr

These forces F1 and F2 do not exist. Centre of mass is not the centre of rotation of the rod.
The centre of rotation of the rod can be found, but lies outside the rod itself.
|dw:1337946684651:dw|
C (centre of mass) is moving down, A is moving horizontally to the right. Centre of rotation I lies at the intersect of the 2 dotted lines, each of them perpendicular to velocity at points C and A.

- anonymous

Ok, I agree that the vertical path of the CM is not the axis of rotation(that was what I stated above), it was not the best usage of the definition of rotational axis. But you are saying that forces F1 and F2 don't exist, how is that so? Let me show you how I found them step by step, and if there is any mistake...?
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- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

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- anonymous

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@Vincent-Lyon.Fr

- Vincent-Lyon.Fr

@ArchiePhysics
You can always resolve a force into components or even many sub-forces: that will have no consequence on N's 2nd law of motion
BUT
if you change the point of action of a force by moving it up or down the rod, then you are changing the torque exerted by this force about a certain point, so you end up with a system of forces that is not equivalent to the one you started with in the first place.
Actually, I do not understand what you are trying to do! The initial question is solved, so what is the 'new' question you are dealing with now? Then I could help.

- anonymous

Oh, I see now.
The thing I'm trying to do is to prove somehow that the center of mass doesn't move horizontally through the fall of the stick(initially, i tried to call it the center of rotation but you've said it's wrong). Because when you say that there are only two vertical forces acting on it and the net force acting on the CM is zero, and thus it does not move horizontally, it isn't that obvious to me))) I'm going to suspect that it isn't right, because my imagination gives me quite complex picture of motion of it, and I'm going crazy on that))) Now the idea of rotational motion through the fall became a bit clearer in my mind, it is thanks to your help guys, and especially thanks to @Vincent-Lyon.Fr thank you very much for the remarks you have made. I really appreciate that!
Cok tesekkur ediyorum, abi!

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