IIT JEE , AIEEE Forum - Physics , Mechanics

atish

Dear sir/madam

I have this small question on rotation from H C Verma. How does an ordinary beam balance work to apply progressively increasing torque against a progressively tlting beam. If you look at it from the energy domain, the change of mech energy due to 1 pan going down and 1 up is same. From force domain, the weight of both pans is same, and due to equal tilt angle in pans, the perpendicular dist is same....leading to equal torque.
How then does the pan going down apply torque to move up ane vice-versa?

I will be very grateful if you could answer my question before 7th april.

Thanking you.
atish.2006@gmail.com">Atish Bhattacharya

atish

Hey guys couldnt any of u answer this??

edison

Actually its not that the pan going down apply torque to move up and vice-versa in case of beam balance.

Anway a good but tricky question the explanation to which is that

When equal masses are suspended on the two sides of beam or physical balance then on tilting the beam (through which the masses are suspended) of the balance teh CENTER OF MASS (CM) of the beam is slightly shifted towards the side which is more closer to the ground or the mass which is more down. Thus the torque due the mass which is above (or towards the upper arm of the beam) increases. And thus it tries to attain the equilibrium position.

moreover the oscillating beam in process of doing so continuously changes the CM and finally is exactly at the center when the equilibrium is restored.

atish

No, I have come across this explanation before, but fail to understand how the CM changes. The beam is rigid, and ideally the CM lies at its centre where it is pivoted. So how should tilting on one side disturb the CM's position?

catch_arnnie

i think here, we are not talking about the CM of the beam, we are looking at the balance as a whole system, so, the CM of the system shifts....

atish

What are you talking about?? The CM of the system plays no role here, and besides it is the last thing to change(if ever) as the centrally symmetrical beams go up or down equally. Please reply with some good answers, this problems been bugging me for ages.

edison

Hi atish,
You are perplexed with the fact that how center of mass is changing.
See here don't get confused with the center which is the point at which the beam is pivoting. This point is not center of mass when beam is tilted.
This u can visualize by slicing/cutting the beam which is tilted or inclined such that the plane of slicing is passing through the center/pivot and vertical to teh ground. Now the portion left to the plane and right to the plane have got different masses. So center of mass shifts towards the portion which is near the ground.
I hope now the problem is more clear. If not ask me again.
You are always welcome.

catch_arnnie

dear mr. edison, how can you say that CM shifts towards the portion which is closer to the ground coz both the parts are goin up & down equally..??

atish

Thank you for showing so much interest in my problem.
From what I understand by looking at your post, you mean to say the CM of whole system is lowered..........due to maybe two possibilities:
1) Due to weight kept on pan
2) Due to lowering of one pan

Well, for case 1, the additional weight is really not a part of the system........say if I just press down on one pan with a force (contact or non-contact), the pan will tilt...apply progressively increasing resistance..........but CM is same.

Case 2 is really lame coz the other pan rises up by the same amount.

atish

Someone pls give a satisfactory answer to this problem......its been a looong time!

abhishekaich

Edison has given a nice explanation to the problem. Let me try to elaborate it a bit.

As pointed out by Edison, when we talk of beam balance the beam is not wieghtless.

To solve the problem for now simply assume that

1. there are no pans attached to the beam

2. The beam is rectangular in shape (with finite width) and hence it centre of gravity is at the intersection of the diagonals.

3. The point at which the beam is suspended is above the CG on the line of symmetry of the beam.

Now try to visualize what happens when the beam tilts to the left. Does the CG remain on the line of symmetry or moves away from it.

You will see it moves to the right of the line of symmetry and hence applies a clockwise torque which restores the position of the beam.

Hope that helps!

phyana

look here atish we u have doubt.............................

step 1. pans empty

step 2. we r placing identical blocks on both the pans.

result :- the beam balance is still. no tilting

step 3. remove one block.

result :- the beam balance tilt to that position where the block is kept

why because the pan with the block is more heavier than the pan without the block.so the torque for the pan with the block is more

along with this process the cm shifts to the pan with the block.

read properly...................
u know torque = force * length.since the pan with the block is more nearer to the cm,the block kept should have that much weight so that it will be able to produce a weight when multiplied wit h the perp dist ,u get a torque ,which is greater than the value(torque) u get when u multiply wt of the other pan and the dis of the other pan from the cm.

that is why beam balance have a specific range where it can measure............

shifting of the cm here just means that.........

assume u have kept a block on one pan.
now cm nearer to that pan.
once u keep a block on the other the cm shifts a bit nearer to the other.

u should realize that the beam balance is not just the rod.........................

if useful then rate meeeeeeeeeeeeeeeeeeeee

atish

Thanks for trying to help

,but its not over yet. abhishekaich, even if I do take all your assumptions to be true, even then the tilting of the beam would cause very litle displacement of the CM which is unlikely to produce so much torque. Anyway, the makers of beam balances use gr8 accuracy(esp the ones used in gold shops) and its very likely that the beam is uniform and pivoted dead centered at its CM/CG. So this theory is not exactly foolproof.
As for your views phyana, I have already written why they are incorrect and I quote here:
"From what I understand by looking at your post, you mean to say the CM of whole system is lowered..........due to maybe two possibilities:
1) Due to weight kept on pan
2) Due to lowering of one pan

Well, for case 1, the additional weight is really not a part of the system........say if I just press down on one pan with a force (contact or non-contact), the pan will tilt...apply progressively increasing resistance..........but CM is same.

Case 2 is really lame coz the other pan rises up by the same amount."

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