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If this is unclear, let me know.

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If this is unclear, let me know.

- #2

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Without external forces, the center of mass will always move with the same velocity, this is momentum conservation.

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The bar is lying flat on a frictionless surface and gravity is neglected. The mass moves at a rate of 1-m/s, so that after the bar has rotated through pi-rad (180-deg), the mass will be on the same side of the bar (if the mass starts on the left and the bar rotates counter-clockwise, the mass will be on the left once again after the bar has finished rotating).

I understand that the conservation of momentum will apply, I am just looking for an equation (or two) to describe the system, including how the bar will continue to rotate around the shifting center of mass.

Thank you, again.

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Then it would help to see the calculations.I am not sure that was the right direction

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I = (m*l

H = I*omega

H1 = pi/3 kg-m

H2 = 0 kg-m

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H1 = pi/3 kg-m

H2 = 0 kg-m

Then:

H1 = 0 kg-m

H2 = -pi/3 kg-m

I tried that out, and it seemed that the rotation direction had to change to get the moments to work out, and I am pretty sure that the bar will keep rotating in the same direction.

Also, I would really like an equation which shows the position of the center-point of the bar as a function of where the center of mass is, i.e. as the center of mass moves from one end to the other, how does the linear position of the center of the bar change. This is in addition to demonstrating that the bar's momentum will continue to carry it around the shifting center of mass (from end to end).

Thank you.

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What is it that will cause the CM to move? Some internal motor?

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the Moment of Inertia will change so the angular velocity will change

Was my understanding correct from my previous post on how momentum will change?

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- #11

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With no external force / couple it can't change. What's the actual model you are working on? We really need a sketch. It may be in your head but it ain't in mine. ;)I agree that it should rotate, but I was hoping for an equation to back it up. The motion is a linear motor, and I am assuming no gravity, drag, or friction.

Was my understanding correct from my previous post on how momentum will change?

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The bar is 1 m with negligible mass. Attached to one end of the bar is a mass of 1 kg. A force of 1 N is applied to the end of the bar opposite the mass. As the bar rotates due to the moment (1 N-m), the mass moves from one end of the bar to the other.

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You do not mention any fixing. Is there any? What direction is the force applied? I just cannot be sure that what I am imagining is what you (think you) are describing. I cannot answer you - and I don't think anyone else can - without a proper diagram with the forces and pivot(?) marked. If you had been in an exam and a question was written as you have written yours, could you even make a start on a solution.

The bar is 1 m with negligible mass. Attached to one end of the bar is a mass of 1 kg. A force of 1 N is applied to the end of the bar opposite the mass. As the bar rotates due to the moment (1 N-m), the mass moves from one end of the bar to the other.

A question properly written is a question half answered. (A PF member - I forget who - often writes that and he is right)

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Here is an image which goes along with my descriptions. Points A and B alternate in position as the mass shifts from end to end of the bar. I am looking to plot the position of point X, the center of the bar.

https://drive.google.com/file/d/0BzHtKK_R-w33TTljSlNoeEZlTDQ/view?usp=sharing [Broken]

https://drive.google.com/file/d/0BzHtKK_R-w33TTljSlNoeEZlTDQ/view?usp=sharing [Broken]

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With a motor, the mass would still not move much and the bar would move "through" the mass, rotating a bit faster when the mass is around the middle of the bar.

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Now it is more clear. If you are dealing with an ideal case then, as the mass approaches the pivot, the angular momentum being unchanged, the angular velocity must increase to infinity. So you have a discontinuity and we're in another ballpark and makes the problem much harder. I can now see why you were asking about the change of direction.

Imo, when the ball goes from one side of the bar to the other, you are in the same sort of situation as when an elastic mass hits an infinitely massive wall. The momentum of the mass will ('actually can') change sign. So the bar, in your case, will change direction. If you change the model slightly from a bar to a disc and you shoot the mass, radially, across the disc, it will hit the disc at a position which depends on the vector sum of rotational v and radial v. Actually, bar with sliding mass or disc and free mass, if they are massless, it doesn't matter I guess because sliding the mass along the wire won't change the momentum situation as the bar will just follow the sliding mass. The external force, to make the angular momentum reverse will come from the small amount of rotation during the transit which alters the tension in the bar and causes an impulsive reaction force from the pivot (a 'bounce').

The above is a bit of jumble but I think it's worth reading as it suggests a different approach from your initial one might get you somewhere. I'll leave it to you to do the actual sums - haha.

Edit: I think there may be a significant difference between the sliding case and the 'fired across' case. Not sure. though.

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- #18

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@sophiecentaur: I don't see how anything would reverse.

If the mass is forced to move along the bar but the rotation of the bar is free, the bar will spin faster and faster. It still has some non-zero mass and the 10kg mass has some finite size and therefore non-zero moment of inertia, so nothing goes to infinity. Move it to the other side, the spinning will get slower again but keep its direction.

If the mass is just pushed once and then free to move on the bar, it will get a bit closer to the center and then move out again, basically following a straight line as the bar just moves according to the motion of the mass.

If you drive both the rotation of the bar and the rotation of the mass, then you fully determine the motion of the system - what is the problem then? You cannot use conservation of momentum or angular momentum if you control the rotation of the bar.

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If there are no external forces acting on the bar, except the initial force which begins the rotation about the center of mass - at the 10 kg mass on the end - and the bar is free to rotate about that center of mass, if the center of mass is constantly being moved from one end of the bar to the other, by conservation of momentum, the bar will continue to rotate about the center of mass, regardless of what end of the bar (A vs B) it is, and the center of the bar (X) will move linearly as the bar rotates about the center of mass with A and B switching places.

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So no pivot? Somehow it is still unclear (at least to me) what exactly the setup is. What is fixed, what is free to change?

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I think this image shows things better:

https://drive.google.com/open?id=0BzHtKK_R-w33M25LUVNGcnkyX2M&authuser=0 [Broken]

https://drive.google.com/open?id=0BzHtKK_R-w33M25LUVNGcnkyX2M&authuser=0 [Broken]

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OK that makes sense. It's the MI of the finite sized mass that takes care of the paradox. I have a feeling that this reduction of rotation speed is along the same lines as the 'yoyo system', which used to be used to cancel the rotation of early satellites that were launched on a spinning rocket.

@sophiecentaur: I don't see how anything would reverse.

If the mass is forced to move along the bar but the rotation of the bar is free, the bar will spin faster and faster. It still has some non-zero mass and the 10kg mass has some finite size and therefore non-zero moment of inertia, so nothing goes to infinity. Move it to the other side, the spinning will get slower again but keep its direction.

If the mass is just pushed once and then free to move on the bar, it will get a bit closer to the center and then move out again, basically following a straight line as the bar just moves according to the motion of the mass.

If you drive both the rotation of the bar and the rotation of the mass, then you fully determine the motion of the system - what is the problem then? You cannot use conservation of momentum or angular momentum if you control the rotation of the bar.

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- #24

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I cannot see the second image because I need "permission". Why not just put your image straight on the PF page? (UPLOAD A FILE is an item under the box that appears when you are composing a post).

I am still confused about the situation - perhaps the diagram would help. But if the bar is on a pivot, you cannot say there are no external forces on it. There will always be a centripetal force from the pivot. Are you actually moving the position of the pivot on the bar or the mass on the bar - or both? Or is X not a pivot?

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Clicking "image" gives me a box to put in a URL, not to upload from my computer. Please try this link:

https://drive.google.com/file/d/0BzHtKK_R-w33M25LUVNGcnkyX2M/view?usp=sharing [Broken]

"X" is the center of the bar, which is what I would like to plot the position of. The bar pivots about the center of mass, which is driven from one end to the other, so that after each 180-degree rotation of the bar, the center of mass is always located on the right end of the bar and the bar (I hope) progresses toward the right due to the conservation of angular momentum which keeps the bar rotating about the (shifting) center of mass.

https://drive.google.com/file/d/0BzHtKK_R-w33M25LUVNGcnkyX2M/view?usp=sharing [Broken]

"X" is the center of the bar, which is what I would like to plot the position of. The bar pivots about the center of mass, which is driven from one end to the other, so that after each 180-degree rotation of the bar, the center of mass is always located on the right end of the bar and the bar (I hope) progresses toward the right due to the conservation of angular momentum which keeps the bar rotating about the (shifting) center of mass.

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- #26

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- the bar is not attached to anything

- the whole system is frictionless

- the mass is fixed at one end of the bar (initially at the right side in the sketch)

- you have a force of 1N acting downwards on the left side for a short time, afterwards the system is free

Are those 4 points correct?

The force will induce a linear motion downwards and a rotation. The size and shape of the 10kg object and the mass of the bar (even if it is very small it is relevant due to its size) are important to determine the angular velocity.

Linear velocity downwards will be given by ##v=\frac{Ft}{10kg}## where F=1N and t is the time the force is applied.

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- #28

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Is my understanding correct?

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You have a radial line of changing length that rotates about an origin and you want to plot the position of the end of this line with respect to the centre of your Mass. In Polar co ordinates it will be (r(t), θ(t)) where r and θ are functions of time, which you can specify. Can you think of anything that formula won't do for you. It is easy to transform into Cartesian coordinates, if you want to show it that way.

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Thank you to everyone who has provided assistance.

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An object that has no mass will not contribute, in any way, to the linear or angular momentum.

Thank you to everyone who has provided assistance.

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