Pulley / conservation of energy problem (should be really easy)

[fvertk]

This problem is bugging the crap out of me. It's the last one I need to do. I swear I got the right answer. Maybe you experts can help? :)

Homework Statement

The system shown in the figure below consists of a light, inextensible cord; light frictionless pulleys, and blocks of equal mass. It is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment when the vertical separation of the blocks is 1.00 m.
http://www.webassign.net/serpop/p7-47.gif

Homework Equations

PEi + KEi = PEf + KEf
KE= (mv^2)/2
PE= mgy

The Attempt at a Solution

1. Judging from what is presented, I assume that the block A moves twice as fast as block B and therefore, it will lower at double the rate that block B will rise.

2. I figured that I would write out my equation to be:
mgy[a] + mgy [initial] = mg(y-2/3)[a] + mg(y+1/3) + (mv^2)/2[a] + (m[v/2]^2)/2

3. Simplified by dividing by m and making initial y to be 0.
0 = g(-2/3) + g(1/3) + (v^2)/2 + ((v/2)^2)/2

4. Input -9.8 for g and simplify more.
-3.26699 = (v^2)/2 + ((v/2)^2)/2

5. Simplify more.
2.556165 = 1.5v

Answer = 1.7 m/s

HOWEVER, that was wrong, and I've checked my math several times. I must be doing something horribly wrong. Kudos and much thanks to you if you can find it, like I said, this has been bugging me for a while.

 

[physixguru]

This problem is bugging the crap out of me. It's the last one I need to do. I swear I got the right answer. Maybe you experts can help? :)

Homework Statement

The system shown in the figure below consists of a light, inextensible cord; light frictionless pulleys, and blocks of equal mass. It is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment when the vertical separation of the blocks is 1.00 m.
http://www.webassign.net/serpop/p7-47.gif

Homework Equations

PEi + KEi = PEf + KEf
KE= (mv^2)/2
PE= mgy

The Attempt at a Solution

1. Judging from what is presented, I assume that the block A moves twice as fast as block B and therefore, it will lower at double the rate that block B will rise.
2. I figured that I would write out my equation to be:
mgy[a] + mgy [initial] = mg(y-2/3)[a] + mg(y+1/3) + (mv^2)/2[a] + (m[v/2]^2)/2

3. Simplified by dividing by m and making initial y to be 0.
0 = g(-2/3) + g(1/3) + (v^2)/2 + ((v/2)^2)/2

4. Input -9.8 for g and simplify more.
-3.26699 = (v^2)/2 + ((v/2)^2)/2

5. Simplify more.
2.556165 = 1.5v

Answer = 1.7 m/s

HOWEVER, that was wrong, and I've checked my math several times. I must be doing something horribly wrong. Kudos and much thanks to you if you can find it, like I said, this has been bugging me for a while.

This statement is a tactical blunder.

 

[fvertk]

Uhm... how so?

I've looked at the system again, and it looks like block A goes down at 2x the rate that block B goes up.

 

[physixguru]

Your reasoning please.

Follow these guidelines for any mechanical problem.:

Draw reference, which is fixed. This reference can be the level of fixed pulley or the ground.
Identify all movable elements like blocks and pulleys (excluding static ones).

Assign variables for the positions of movable elements from the chosen reference.

Write down constraint relations for the positions of the elements. Usually, total length of the string is the “constraint” that we need to make use for writing relation for the positions.

Differentiate the relation for positions once to get relation of velocities and twice to get relation of accelerations.

 

[alphysicist]

Hi fvertk,

I think you have some algebra errors:

4. Input -9.8 for g and simplify more.
-3.26699 = (v^2)/2 + ((v/2)^2)/2

The left side is negative, yet the right side is a positive number. The next step also does not seem to follow this one.

 

[fvertk]

Well, when looking at the problem, it appeared that way initially. I looked up other responses to it, and that's what someone here had to say about it too:
https://www.physicsforums.com/showthread.php?t=95263
(read Doc Al's post)

Now that I look at it again, I guess they actually ARE rising and falling at the same rate. (Is that what you're hinting at?) However, when I put that into the equation, all I get is 0. It can't be solved that way apparently. So I tried it for one block, doing mgy= mgy + mv^2/2 (where y(final) is -.5 and y(initial) is 0) and I got 3.13, which was still the wrong answer.

PS: To be honest with you, if the blocks are the same mass, I don't understand why they are moving in the first place. Shouldn't it be completely stationary?

 

[alphysicist]

Well, when looking at the problem, it appeared that way initially. I looked up other responses to it, and that's what someone here had to say about it too:
https://www.physicsforums.com/showthread.php?t=95263
(read Doc Al's post)

Now that I look at it again, I guess they actually ARE rising and falling at the same rate. (Is that what you're hinting at?) However, when I put that into the equation, all I get is 0. It can't be solved that way apparently. So I tried it for one block, doing mgy= mgy + mv^2/2 (where y(final) is -.5 and y(initial) is 0) and I got 3.13, which was still the wrong answer.

PS: To be honest with you, if the blocks are the same mass, I don't understand why they are moving in the first place. Shouldn't it be completely stationary?

I think your original equation was okay; you just had the algebra errors I mentioned in my post. If you plug your answer back into what you have for equation 3 you'll find it doesn't work.

 

[Doc Al]

I've looked at the system again, and it looks like block A goes down at 2x the rate that block B goes up.

Sounds right to me. (But check the errors that alphysicist pointed out.)

 

[fvertk]

Hi fvertk,

I think you have some algebra errors:

4. Input -9.8 for g and simplify more.
-3.26699 = (v^2)/2 + ((v/2)^2)/2

The left side is negative, yet the right side is a positive number. The next step also does not seem to follow this one.

Right, I did change that to positive in order to get the square root. It shouldn't matter what sign it is since I'm trying to find the speed, right? And at that point, I'm no longer adding/subtracting, but multiplying.

What I did to get to step five was multiply 2.36699 by 2, then find the square root of that.

 

[alphysicist]

You have:

3.26699 = (v^2)/2 + ((v/2)^2)/2

So when you multiply by 2, you get:

6.53398 = v^2 + v^2 /4

What do you get when you take the square root of both sides?

 

[fvertk]

Okay, I see what you're saying alphysicist. Looks like I gave it the wrong sign in step 4 in my addition from step 3. But that should be okay, since I changed it anyway... I'm not sure that's the problem then.

 

[fvertk]

You have:

3.26699 = (v^2)/2 + ((v/2)^2)/2

So when you multiply by 2, you get:

6.53398 = v^2 + v^2 /4

What do you get when you take the square root of both sides?

Why is the latter part of that v^2/4? How did you eliminate the fraction beneath the exponent? What I did was simplify the 1/2 to be from this:
3.26699 = (v^2)/2 + ((v/2)^2)/2

To this:
3.26699 = 1/2((v^2) + ((v/2)^2))

And then multiplied both sides by two, getting me this:

6.53398 = v^2 + (v/2)^2

Then after doing the square root, I get:

2.556 = v + v/2
Maybe I'm wrong though... perhaps there's an algebra trick I haven't learned yet? :/

 

[alphysicist]

Because in the last term,

(v/2)^2

means squaring both numerator and denominator.

$$\left(\frac{v}{2}\right)^2 = \frac{v^2}{4}$$

 

[fvertk]

Oh yeah, duh. *smacks head* Although, shouldn't my technique work just as well? Let me try yours though, hold on...

 

[alphysicist]

Why is the latter part of that v^2/4? How did you eliminate the fraction beneath the exponent? What I did was simplify the 1/2 to be from this:
3.26699 = (v^2)/2 + ((v/2)^2)/2

To this:
3.26699 = 1/2((v^2) + ((v/2)^2))

And then multiplied both sides by two, getting me this:

6.53398 = v^2 + (v/2)^2

Then after doing the square root, I get:

2.556 = v + v/2
Maybe I'm wrong though... perhaps there's an algebra trick I haven't learned yet? :/

No, it doesn't work that way. For example,

$$\sqrt{3^2 + 4^2} = 5 \mbox{ (not 7)}$$

First combine the two terms on the right side. v^2 + (1/4) v^2 equals what?

Then you might want to get v^2 by itself on the right side and then take the square root.

 

[fvertk]

I see... I must have really botched that square root then by doing two under one. Strange, I guess I had some misconception that that actually worked. Nice catch, that actually gave me the right answer. Thanks alphysicist.

By the way, physixguru, I'm not sure what you were talking about above then. Seems like it did go 2x faster, but thanks for helping anyway, I'm just confused.

 

[alphysicist]

By the way, in an earlier post you mentioned that you did not understand why it was moving in the first place. Does that make sense now?

 

[fvertk]

Yes, in the case that block A has a faster velocity downward, it does. When I doubted that initial assessment of the system, thinking that each block has equal velocity, that's when I got confused.

 

[rl.bhat]

While solving this problem you have taken the acceleration of the blocks as g, but it is not g. Since the single rope is passed through two pullyes, the tension T in each segment of the rope must be the same. Since block B is pulled up by 2T and block A by T, block A moves in the downward direction. The acceleration of block A is (mg - T)/m. And acceleration of block B is (2T - mg)/m. Solve for T and find the value of acceleration. Then apply the conservation law of energy for the block A or kinamatics equation v^2 = vo^2 + 2ah. As you have said the disatance moved by A is (2/3)m

 

[alphysicist]

Hi rl.bhat,

While solving this problem you have taken the acceleration of the blocks as g, but it is not g. Since the single rope is passed through two pullyes, the tension T in each segment of the rope must be the same. Since block B is pulled up by 2T and block A by T, block A moves in the downward direction. The acceleration of block A is (mg - T)/m. And acceleration of block B is (2T - mg)/m. Solve for T and find the value of acceleration. Then apply the conservation law of energy for the block A or kinamatics equation v^2 = vo^2 + 2ah. As you have said the disatance moved by A is (2/3)m

I'm not sure what you are referring to; the acceleration was not calculated at all in the problem, so I don't think it was assumed to be g. Here an energy approach was used which gave the correct answer (after some algebraic answers were corrected).

The effect of the ropes was handled by constraining the speed and distances of one mass to be twice that of the other.

 

[physixguru]

I see... I must have really botched that square root then by doing two under one. Strange, I guess I had some misconception that that actually worked. Nice catch, that actually gave me the right answer. Thanks alphysicist.

By the way, physixguru, I'm not sure what you were talking about above then. Seems like it did go 2x faster, but thanks for helping anyway, I'm just confused.

Dear fvertk, i told that the statement you made was a tactical blunder.It meant that it did not go in accordance with what you were showing mathematically.Either you were theorotically wrong or mathematically wrong.I told you so to prompt you to tell what you actually thought of your approach.The most important thing about solving any problem is the correct reasoning.I thought you were pretty confused with the maths and did not abide by the theory you knew.Had i directly pointed you out the error maybe you wud not have thought over it again.Process is more important than answer.

 

[Doc Al]

Dear fvertk, i told that the statement you made was a tactical blunder.It meant that it did not go in accordance with what you were showing mathematically.Either you were theorotically wrong or mathematically wrong.I told you so to prompt you to tell what you actually thought of your approach.The most important thing about solving any problem is the correct reasoning.I thought you were pretty confused with the maths and did not abide by the theory you knew.Had i directly pointed you out the error maybe you wud not have thought over it again.Process is more important than answer.

Why did you highlight fvertk's sentence "I assume that the block A moves twice as fast as block B and therefore, it will lower at double the rate that block B will rise." and refer to it as a "tactical blunder"? That highlighted statement is correct.

 

[physixguru]

Why did you highlight fvertk's sentence "I assume that the block A moves twice as fast as block B and therefore, it will lower at double the rate that block B will rise." and refer to it as a "tactical blunder"? That highlighted statement is correct.

Sorry to argue sir, my point is that the above statement could have been expressed in a better way.As far as i know, my reasoning would have been different.His applied tactics in the procedure were wrong.Blunder meant that he was contradicting his own mathematical solution.

 

[alphysicist]

Hi physixguru,

Sorry to argue sir, my point is that the above statement could have been expressed in a better way.As far as i know, my reasoning would have been different.His applied tactics in the procedure were wrong.Blunder meant that he was contradicting his own mathematical solution.

If you don't mind looking at this a bit further...

I don't see anything wrong with the applied tactics in his solution. Fvertk used an energy approach, and needed the relative speeds and relative displacements of the two objects. To me, the sentence you referred to handles both of those aspects clearly and was the correct approach.

The only errors I saw were two purely mathematical errors, done after the energy equation was written down correctly. One was a sign error (which was self-corrected), and the other error was assuming $\sqrt{a^2 + b^2 } \rightarrow a+b$.

From what I can see, fvertk had done the physics part of this problem correctly (the theory and the data and constraints from the problem were all put together in the right way).

 

[physixguru]

Hi physixguru,



If you don't mind looking at this a bit further...

I don't see anything wrong with the applied tactics in his solution. Fvertk used an energy approach, and needed the relative speeds and relative displacements of the two objects. To me, the sentence you referred to handles both of those aspects clearly and was the correct approach.

The only errors I saw were two purely mathematical errors, done after the energy equation was written down correctly. One was a sign error (which was self-corrected), and the other error was assuming $\sqrt{a^2 + b^2 } \rightarrow a+b$.

From what I can see, fvertk had done the physics part of this problem correctly (the theory and the data and constraints from the problem were all put together in the right way).

Sorry again sir,but i wud like to contradict you.The overall approach was not wrong but the he said that the basis of his solving the problem was the constraint relation he got ,which was evident from his statement which i highlighted.According to me, a question always needs multidimensional thinking and not a one dimeansional approach.I would have begun my problem on a different note.Even a toddler knows that he has to start solving a mechanics problem using mass-acceleration relations but the people who really analyse the problem , go into the depth of it rather than just applying the bookish knowledge.

sorry for arguing , anyways.You all are much senior, and ofcourse know much more.