Beads on a string on a peg

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In summary, the problem involves a massless string with three identical beads attached to it hanging on a frictionless peg. The string slides over the peg and falls to the ground. Given a distance of 55 cm, the problem asks for the speed of the lowest bead when the upper bead passes over the peg. By considering the changes in potential energy and kinetic energy for all three beads in the system, it can be determined that the final kinetic energy is equal to the initial potential energy, resulting in a speed of 3.28 m/s for the lowest bead at the given moment.
  • #1
omc1
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Homework Statement

A massless string having three beads of identical mass attached to it is placed so that it hangs at rest on a thin frictionless peg, as shown. Of course, the string then slides over the peg and falls to the ground. If h = 55 cm, what is the speed of the lowest bead at the instant when the upper bead passes over the peg

Homework Equations

ke=1/2mv^2 w=fd

The Attempt at a Solution

Δke=fd 1/2mv^2 final =0 1/2mv^2 (initial)=fd
d=55cm i got 3.28m/s but that's not right
 
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  • #2
A diagram or at least a better description of the setup would be helpful.
 
  • #3
three beads are on a string and and laying over a peg protruding from the wall...so that beads one and two are even (bead one hangs on the left of peg and bead two on the right) and the third bead hangs down on right just farther down. h = the distance from the peg to the first (or second bead) and also the distance from first bead to third bead so H =2h...
 
  • #4
So, something like this?

attachment.php?attachmentid=51711&stc=1&d=1349819561.gif


Can you provide a bit more detail about the calculation you performed? What was your reasoning?
 

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  • #5
yes!
 
  • #6
i figured if i found the speed of the first bead going over the peg it would be the same as the thrid bead. so i made the first bead start at 0 and end at 55 cm so i did 1/2mv^2=mgd and got 3.28m/s
 
  • #7
omc1 said:
i figured if i found the speed of the first bead going over the peg it would be the same as the thrid bead. so i made the first bead start at 0 and end at 55 cm so i did 1/2mv^2=mgd and got 3.28m/s

Okay, but you do realize that a bead cannot spontaneously gain energy by falling upwards, right? Only something falling in the gravitational potential can gain kinetic energy, while something rising against gravity will lose it.

So, identify the source of the energy gain and offset it against any losses... :wink:
 
  • #8
the total distence and the total mass of 2 and 3 beads are what's giving bead one the potential to move upward so pe=m(sub total)g2h ??
 
  • #9
Be sure to account for ALL the changes in PE.
 
  • #10
well wouldn't the final pe just be zero bc it hits the ground??
 
  • #11
omc1 said:
well wouldn't the final pe just be zero bc it hits the ground??

There is no mention of the ground anywhere in the problem statement.

Some beads rise, some beads fall... what's the total change in PE?
 
  • #12
pe=mgh(final)-mgh(initial) so if bead one starts out at zero and ends at 55cm then pe=mgh(final)...
 
  • #13
As I said, some parts are rising and some parts are falling. The parts that are rising are going to to gain PE. The ones that are falling are going to lose PE. Do the sums!
 
  • #14
ok but i don't see where ur going with this...iam just confused n suck at physics
 
  • #15
omc1 said:
ok but i don't see where ur going with this...iam just confused n suck at physics

It takes energy to raise a mass in a gravitational field. You also get energy out if you lower a mass in a gravitational field. Energy is conserved. So if some of the mass is rising and some of the mass is falling, what's the net change in energy that results from all the movements? You can't just look at the change in height of one bead. You must take ALL of them and their motions into account.
 
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  • #16
so the total energy for the system would be the sum of pe and ke ...?
 
  • #17
omc1 said:
so the total energy for the system would be the sum of pe and ke ...?

Yes, and the sum is a constant (conservation of energy). But you get to choose your zero reference points for convenience. The system starts with zero KE (no motion). You could also assign a zero PE reference individually to the masses at their initial locations and just sum their changes in PE for a total change in PE.
 
  • #18
ok i seee now thanks so much!
 
  • #19
omc1 said:
ok i seee now thanks so much!

Glad to help!
 

1. What is the purpose of "Beads on a string on a peg" in science?

The "Beads on a string on a peg" model is used in science to demonstrate the concept of molecular motion and how it relates to the three states of matter - solid, liquid, and gas. It can help scientists understand the behavior and properties of different substances at the molecular level.

2. How does the "Beads on a string on a peg" model work?

The model consists of a string, representing the bonds between molecules, and beads, representing the molecules themselves. These beads are attached to a peg, which serves as a fixed point. By moving the peg, the string and beads can be manipulated to simulate the movement of molecules in different states of matter.

3. What can "Beads on a string on a peg" teach us about states of matter?

The model can help us understand that in a solid state, the molecules are tightly packed and have limited movement, while in a liquid state, the molecules have more freedom to move and can slide past each other. In a gas state, the molecules are spread out and have the most freedom to move.

4. Can "Beads on a string on a peg" be used to study other scientific concepts?

Yes, the model can also be used to demonstrate other concepts such as diffusion, surface tension, and the effect of temperature on molecular motion. It can also be adapted to represent different types of molecules, such as polymers or proteins.

5. How is "Beads on a string on a peg" relevant to real-world applications?

The model can be used to better understand and predict the behavior of different substances in various conditions, which can be applied in fields such as material science, chemistry, and engineering. It can also be used in educational settings to help students visualize and grasp scientific concepts.

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