Rotational and Translational equilibrium

[happyfellow]

I've recently missed a whole slew of questions about what happens when we disrupt the rotational equilibrium of a beam on a fulcrum. In particular, what happens to translational equilibrium when our rotational equilibrium is disrupted?

I am always able to pick out when rotational equil. is disrupted by using rxF = rxF, but I don't know what this means for translational equilibrium. Any insight would be appreciated.

-happy

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[IntelInside]

I've recently missed a whole slew of questions about what happens when we disrupt the rotational equilibrium of a beam on a fulcrum. In particular, what happens to translational equilibrium when our rotational equilibrium is disrupted?

I am always able to pick out when rotational equil. is disrupted by using rxF = rxF, but I don't know what this means for translational equilibrium. Any insight would be appreciated.

-happy

can you give us some of the questions so we can help you work through them?

 

[happyfellow]

Here's one that I missed from a passage about a laboratory balance. Initially the rod is balanced.

If one of the weights on the balance is moved from 1cm to 5cm:

A) the rod is no longer in translational or rotational equilibrium
B) the rod is in translational equilibrium, not not in rotational equilibrium
C) the rod is in rotational equilibrium, but not in translational equilbrium
D) the rod is still in both translational and rotational equilibrium

Answer is A. I know that moving this particular weight on the rod will offset rotational equilibrium (using r X F = r X F) but I don't understand why the translational equilbrium is disrupted as well.

 

[IntelInside]

Here's one that I missed from a passage about a laboratory balance. Initially the rod is balanced.

If one of the weights on the balance is moved from 1cm to 5cm:

A) the rod is no longer in translational or rotational equilibrium
B) the rod is in translational equilibrium, not not in rotational equilibrium
C) the rod is in rotational equilibrium, but not in translational equilbrium
D) the rod is still in both translational and rotational equilibrium

Answer is A. I know that moving this particular weight on the rod will offset rotational equilibrium (using r X F = r X F) but I don't understand why the translational equilbrium is disrupted as well.

Interesting....I dont see how linear motion in either the x or the y direction would be affected since all we are affecting by moving the weights is conveying an angular acceleration to it, not a linear accel.

 

[UIUCstudent]

Bump. I'm having this trouble too in a similar question in one of the BR physic passage. Can someone enlighten me?

 

[BerkReviewTeach]

Bump. I'm having this trouble too in a similar question in one of the BR physic passage. Can someone enlighten me?

If the center of mass moves far enough to one side or the other, there is the chance it will no longer be offset by the normal force of the base, and it falls. Think about what happens if you balance a pen on your finger. If you added a small mass (like a piece of gum) to one end of the pen, it would both rotate and fall. It rotates because of the torque imbalance and it falls because the weight is no longer being supported by your finger. Falling from rest implies it has a linear force acting on it, so it can't be in translational equilibrium.

 

[sillyether]

If the center of mass moves far enough to one side or the other, there is the chance it will no longer be offset by the normal force of the base, and it falls. Think about what happens if you balance a pen on your finger. If you added a small mass (like a piece of gum) to one end of the pen, it would both rotate and fall. It rotates because of the torque imbalance and it falls because the weight is no longer being supported by your finger. Falling from rest implies it has a linear force acting on it, so it can't be in translational equilibrium.

In that case, would it be fair to say that if something is not in rotational equilibrium, then it necessarily can't be in translational equilibrium either?

 

[sillyether]

In other words, there are 2 questions I would like answered:
1. Can translational equilibrium exist without rotational equilibrium?

2. If a pivot is not placed at the center of mass of a meter stick (that weighs, say 1kg), how much of the weight of the weight of the meter stick is supported by the pivot? I know this depends on where the pivot is, but is it always the case that as long as it's not at the center of mass, it'll support LESS than the total weight of the meter stick?

Thanks! This would clear up a lot of concepts for me!

 

[milski]

In other words, there are 2 questions I would like answered:
1. Can translational equilibrium exist without rotational equilibrium?

Yes. A seesaw with only one person on one side would be example of that - it will rotate but will not (hopefully) fall off from its hinge in the middle. Generally you need some sort of "attachment" at the pivot point to have the appropriate normal force to maintain the translational equilibrium.

2. If a pivot is not placed at the center of mass of a meter stick (that weighs, say 1kg), how much of the weight of the weight of the meter stick is supported by the pivot? I know this depends on where the pivot is, but is it always the case that as long as it's not at the center of mass, it'll support LESS than the total weight of the meter stick?

Thanks! This would clear up a lot of concepts for me!

That depends - are there any other support points, generating normal force on the meter stick? If the pivot is the only support point, the normal will be the same and will not change if you move the stick left or right.

 

[SaintJude]

If the center of mass moves far enough to one side or the other, there is the chance it will no longer be offset by the normal force of the base, and it falls. Think about what happens if you balance a pen on your finger. If you added a small mass (like a piece of gum) to one end of the pen, it would both rotate and fall. It rotates because of the torque imbalance and it falls because the weight is no longer being supported by your finger. Falling from rest implies it has a linear force acting on it, so it can't be in translational equilibrium.

Yes,good concept teaching BUT nowhere in the question stem does it indicate the balance suddenly falls or tips over. Or is it stated in the passage?

 

[Lunasly]

I'd also like more insight. I got the same problem incorrect. How do you know that the centre of mass changed enough for the system to move out of translational equilibrium?

My impression of this concept after reading the answer is that since the fulcrum is not physically attached to the rod (I assume), a change in the centre of the mass can cause the system to move out of equilibrium. As the system rotates (clockwise in this example), the centre of mass shifts to the right and so now the net force in the downward direction exceeds the net force in the upward direction (normal force from the fulcrum).

Anyone else have other ways of conceptualizing this problem?

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Edit: I think I have a solution. Hopefully someone can correct me. I think question 12 - 14 for this passage all relate to different scenarios, not necessarily the scenario presented in the passage.

For instance, if you want a system to be in equilibrium such as is the case of two masses on a bar balanced on a fulcrum, then both net translational and net rotational forces must = 0. If one of them changes, then you can no longer be in equilibrium.

Now, if we move away from this scenario and talk about another unrelated scenario, then you can have no net force, but still have a net torque. Likewise, you could have no net torque and a net force. A net torque = 0 tells us that we not angular acceleration. A net force = 0 means we have no translational acceleration. However, in scenarios unrelated the one described in this specific passage (two masses on a bar balanced on a fulcrum), there is no reason why we can't have a net force and no net torque (think of an object in free-fall). It has a net force in the downward direction due to gravity, but no net torque because it does not rotate. Likewise, if we think of the Price is Right wheel, that has a net torque (i.e., a net angular acceleration) due to the force we apply to it, but it is in translational equilibrium since there is no net force (it doesn't fall since it is physically attached).

In summary (and someone please correct me if I am wrong), I think that this passage is a very specific scenario where both net rotational and translational forces must = 0 for the system to be in equilibrium as the centre of mass can change. In the scenarios I described above, I don't think we can change the centre of mass.

Hopefully someone can double check this.