Rotation of free solid object in vacuum, location of axis of rotation


by Holali
Tags: axis, force, motion, rotation, vacuum
Holali
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#1
May6-12, 06:35 AM
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Hi,

I would be happy if you could solve and explain following problem:
Imagine some solid long object (wooden plank for example) being just like that in vacuum, free, without being fixed to any point of space.
Then, imagine only one forse pushing this object in a certain point of object, perpendicular to object's length. This force remains the same during time.
What I want to know is, how would such object move in space. I guess it will experience both translational and rotational motions. But where will be the axis of rotation? And how would speed of translation and rotation depend on strength of the force, on the mass of the plank and on the operation point of the force..

image for visualization
h*t*t*p://imageshack.us/photo/my-images/52/physicsimg.png/

Sorry for my bad english at some points, hope you understand the problem anyway.
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Doc Al
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#2
May6-12, 07:15 AM
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The force produces both a translational acceleration of the center of mass (F = ma) and a rotational acceleration about the center of mass (Torque = Iα). The motion of the plank will be the sum of those two effects.
vin300
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#3
May6-12, 11:11 AM
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All of the energy supplied to the plank does not translate into a single form. Some of it becomes its angular kinetic energy and the rest becomes translational. I don't know what the ratio will be.I had the exact same doubt.

vin300
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#4
May6-12, 11:36 AM
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Rotation of free solid object in vacuum, location of axis of rotation


From the experiment I conducted myself with a pencil I concluded that the ratio of the rotational and translational energies increases as the force is applied farther from the centre of mass and is zero when it is applied at the centre of mass.
Holali
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#5
May6-12, 12:38 PM
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Thanks for your replies, but there are still unsolved things:
How fast will the plank rotate and around what point? (are you sure it will be around centre of the mass?)
And what will be the speed of translational motion?
Doc Al
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#6
May6-12, 01:40 PM
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Quote Quote by Holali View Post
Thanks for your replies, but there are still unsolved things:
How fast will the plank rotate and around what point?
You'll figure out the instantaneous axis of rotation by adding the translational and rotational motion (about the center of mass).
(are you sure it will be around centre of the mass?)
Never said it would. But you can compute the rotation about the center of mass using Torque = Iα, then add it to the translational motion.
And what will be the speed of translational motion?
The acceleration of the center of mass is given by F = ma.
vin300
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#7
May6-12, 02:05 PM
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Quote Quote by Doc Al View Post
You'll figure out the instantaneous axis of rotation by adding the translational and rotational motion (about the center of mass).
What does this mean? The body always rotates about its centre of mass because in this state, the centrifugal forces will be balanced in the plane of rotation.The centrifugal force is proportional to the product of mass and radius, for the same angular velocity.
EDIT: The above occurs after the force has been removed. While the force is still in action, the instantaneous centre of rotation will vary from one instant to another.

The acceleration of the center of mass is given by F = ma.
This is only true if the only thing the force does is causing linear acceleration. Here the force also causes rotation.It breaks energy conservation.
Doc Al
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#8
May6-12, 02:25 PM
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Quote Quote by vin300 View Post
What does this mean? The body always rotates about its centre of mass because in this state, the centrifugal forces will be balanced in the plane of rotation.The centrifugal force is proportional to the product of mass and radius, for the same angular velocity.
Huh? Why in the world are you introducing centrifugal forces to this discussion?
EDIT: The above occurs after the force has been removed. While the force is still in action, the instantaneous centre of rotation will vary from one instant to another.
We are talking about a force acting. And even without a force, if something translates as well as rotates--as is the case here--the instantaneous axis of rotation will vary.
This is only true if the only thing the force does is causing linear acceleration. Here the force also causes rotation.It breaks energy conservation.
This is incorrect. Newton's law works just fine, as does energy conservation.
cepheid
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May6-12, 02:26 PM
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Quote Quote by vin300 View Post
This is only true if the only thing the force does is causing linear acceleration. Here the force also causes rotation.It breaks energy conservation.
This is false. F = ma holds true in this situation like in any other. The energy for the translational motion comes from the work done by the force [itex] W = \int F dx [/itex] and the energy for the rotational motion comes from the work done by the torque [itex] W = \int \tau d\theta [/itex]. These integrals give you the expressions for translational and rotational kinetic energy respectively
vin300
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#10
May6-12, 03:14 PM
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This is only true if the only thing the force does is causing linear acceleration. Here the force also causes rotation.It breaks energy conservation.
I realise I was wrong about this, after some pondering.
vin300
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#11
May6-12, 03:36 PM
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Quote Quote by Doc Al View Post
We are talking about a force acting. And even without a force, if something translates as well as rotates--as is the case here--the instantaneous axis of rotation will vary.
Yes but it will always pass through the centre of mass, won't it?
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#12
May6-12, 04:23 PM
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Quote Quote by vin300 View Post
Yes but it will always pass through the centre of mass, won't it?
No..
vin300
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#13
May6-12, 10:07 PM
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Right.I got it.
Holali
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#14
May7-12, 04:19 PM
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Ok, thanks for all posts, its become quite big discussion here. Can somone make some conclusion now?
The best way to understand this problem is via an example.
If the plank has the mass m=2kg and length 12m, the force pushing it is in distance d=3 meters from the centre of the mass and F=4N, and its acting for 3 seconds, whatt will be the result? I mean, what will be the angular speed and speed of translation? And what direction?
I'm sorry if I'm too annoying with so many questions. I would really like to understand this problem deeply.
Thanks anyway
cepheid
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May7-12, 07:21 PM
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Quote Quote by Holali View Post
Ok, thanks for all posts, its become quite big discussion here. Can somone make some conclusion now?
The best way to understand this problem is via an example.
If the plank has the mass m=2kg and length 12m, the force pushing it is in distance d=3 meters from the centre of the mass and F=4N, and its acting for 3 seconds, whatt will be the result? I mean, what will be the angular speed and speed of translation? And what direction?
I'm sorry if I'm too annoying with so many questions. I would really like to understand this problem deeply.
Thanks anyway
I kind of thought that your questions had been answered already. In any case, the problem you give above is not perfectly well-defined, is it? Does the force always act at a specific point on the plank that happens to be located 3 m from the centre of mass? If so, then as the plank begins to spin, the direction of the force (and hence the direction of the acceleration) changes with time. That makes it more complicated to compute the speed vs. time. If we can assume that this effect is small enough (over 3 seconds) be ignored, and we pretend that the force is constant in both magnitude and direction, then the first part of your question is easy: the acceleration will be 2 (m/s)/s and hence the velocity after 3 seconds (assuming the plank starts from rest) will be 6 m/s in the direction in which the force is being applied.

The torque around the centre of mass is easy enough to calculate: 12 Nm assuming the force is applied normal to the surface of the plank. To get the angular acceleration, we need more information: what is the shape of the plank? You have only given one dimension. If this is the only dimension that is important, and the others are negligible, then the object is more of a "thin rod" than a plank. For it to be a plank, I would think that at least one other dimension would be non-negligible i.e. you'd need a width as well as a length. In any case, we need the exact shape of the plank in order to compute its moment of inertia and hence the angular acceleration around the centre of mass, given the above torque.
Holali
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#16
May10-12, 08:31 AM
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Well, it can be also thin rod. I originaly meant a long prism with square base. But rod is fine too. Does it matter a lot if it has square or circular base? (if the volume and mass are the same). We can use anything long with known moment of inertia.
And yes, the force is all the time acting perpendicular to the length of object, and alway 3m away from the centre of the mass. (imagine for example small rocket engine with constant force attached and fixed to it). So, the direction of the force is not constant during time, but its constantly perpendicular to the length of the object.
Hope you understand my strange description.
Is this everything you need to know to solve this problem?
I wat to know what will the force cause.
A.T.
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#17
May10-12, 09:17 AM
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Quote Quote by Holali View Post
(imagine for example small rocket engine with constant force attached and fixed to it). So, the direction of the force is not constant during time, but its constantly perpendicular to the length of the object.
Hope you understand my strange description.
Is this everything you need to know to solve this problem?
I wat to know what will the force cause.
First you integrate the angular acceleration (from torque and inertial moment) around the COM twice, to get the orientation as function of time (and the final angular velocity). Then you use the orientation function and thrust force to get the linear acceleration over time, which integrated once gives you the final linear velocity of the COM.


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