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Rotation of free solid object in vacuum, location of axis of rotation 
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#1
May612, 06:35 AM

P: 9

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/myimages/52/physicsimg.png/ Sorry for my bad english at some points, hope you understand the problem anyway. 


#2
May612, 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.



#3
May612, 11:11 AM

P: 513

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.



#4
May612, 11:36 AM

P: 513

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.



#5
May612, 12:38 PM

P: 9

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? 


#6
May612, 01:40 PM

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#7
May612, 02:05 PM

P: 513

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. 


#8
May612, 02:25 PM

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#9
May612, 02:26 PM

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#10
May612, 03:14 PM

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#11
May612, 03:36 PM

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#13
May612, 10:07 PM

P: 513

Right.I got it.



#14
May712, 04:19 PM

P: 9

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 


#15
May712, 07:21 PM

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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 nonnegligible 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. 


#16
May1012, 08:31 AM

P: 9

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. 


#17
May1012, 09:17 AM

P: 4,212




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