# Homework Help: Initial acceleration of block when string is cut

1. Mar 14, 2014

### Tanya Sharma

1. The problem statement, all variables and given/known data

In figure shown, pulleys are ideal m1 > 2 m2. Initially the system is in equilibrium and string connecting m2 to rigid support below is cut. Find the initial acceleration of m2?

2. Relevant equations

3. The attempt at a solution

In equilibrium,

2T1 = T2
For block m1 , T2 = m1g
For block m2, T1 = m2g + T3

When the string is cut,let the initial acceleration be 'a'

2T1 = T2
For block m1 , m1g - T2 = m1a/2
For block m2 ,T1 - m2g = m2a

This gives , a=2(m1-2m2)g/(m1+4m2) ,which is incorrect answer .

I would be grateful if somebody could help me with the problem .

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2. Mar 15, 2014

### Simon Bridge

You want the acceleration of m2 right after the string is cut.

Concept:
Instantly after it is cut: T1 is the only force acting on m2 - at the same value as immediately before the string was cut.

3. Mar 15, 2014

### paisiello2

A couple of things:

-is m1 = 2*m2 or is m1 > 2*m2?

-one practical thing to keep in mind is that the purpose of a pulley system is to be able to use a small force over a large distance to lift a heavy force over a small distance

-that being said, the distance that m2 travels is twice the distance that m1 travels and therefore the acceleration is twice as well (if you think about the total length of the rope being constant then you should be able to convince yourself that this is true)

I think that is the only thing you were missing.

4. Mar 15, 2014

### Tanya Sharma

That is not right .T1 is not the only force acting on m2 after the string is cut.

Last edited: Mar 15, 2014
5. Mar 15, 2014

### paisiello2

Sorry, I missed that you did that. I won't bother you anymore.

6. Mar 15, 2014

### haruspex

No, there's still gravity of course, but Simon's point is that you are ignoring the presence of the spring. That ensures T1 does not immediately diminish when the string is cut.

7. Mar 15, 2014

### Tanya Sharma

Hi haruspex

Thanks for chipping in ...

1)Does that mean the initial acceleration of m1 is zero ?

2)Does that mean if instead of spring there was a string ,then the tension T1 would diminish immediately ?

3) How is initial acceleration i.e at t=0 different from acceleration at any time t>0 ?

8. Mar 15, 2014

### haruspex

You're welcome. had limited net access for a while.
No, it means that its aceleration is not a/2. Since the spring will contract, the string between it and m1 will accelerate upwards.
Yes
As the spring contracts, it will exert less force.

9. Mar 15, 2014

### Tanya Sharma

Okay...but you were missed :)

Please reconsider this .I think the initial acceleration of m1 should be zero ,as forces on m1 do not change .If it is not zero then ,it would not be possible to calculate initial acceleration of m2 .

A few more doubts

1) In the setup,because of the presence of spring ,the acceleration of the two blocks will be independent of each other i.e no relationship between them . But if it were string ,then a2 = 2a1 at all times .Is it correct ?

2) T1 = Kx at all times .i.e tension would vary . Is that so ?

3) I am having difficulty analyzing why T1 would diminish immediately if there were a a string instead of a spring ? If there were a string ,then ,why would block m2 move upwards even though the only force would be m2g downwards ?

Last edited: Mar 15, 2014
10. Mar 15, 2014

### haruspex

Yes, you're right. This occurred to me later but I'd lost access again.
I wouldn't say no relationship, but certainly a different one.
Yes.
I said diminish, not vanish. Before the cut, T1 may considerably exceed m2g. After the cut it is reduced but still exceeds m2g (because m1 exceeds 2m2).

11. Mar 15, 2014

### Tanya Sharma

So you are getting flashes of internet connectivity . Hope it improves soon.

You are right .I misinterpreted the statement.

Thanks !

12. Mar 16, 2014

### ehild

As the length of the string+spring is not constant, a1 is not equal to a2/2.

I do not like such questions: "what would be the initial acceleration when something changes suddenly". In case of a sudden change such factors count what we usually neglect. The mass of the pulleys and the mass of the string. The elasticity of the string.

In this problem, the spring exerts some force on the string and it takes time till the length of the spring and the spring force change. As there is force at one end of the string there must be the same, but opposite force at the other end, if the pulleys and string are really massless. That is what the problem-writer probably assumed.

The tension is built up in the string as if it was also a spring. Because of the load, its molecules depart slightly from the equilibrium position. But the "spring constant" is very high so the new tension is built up in a very short time. And also the stretching of the string is very small so we can take its length constant.

There is time needed to get the stationary forces and the "stationary" acceleration in both cases. The difference between the case with and without the spring is the duration of this time intervals.

That the tension in a spring does not disappear at once is shown nicely with slinky experiments. See:

ehild

Last edited by a moderator: Sep 25, 2014
13. Mar 16, 2014

### Tanya Sharma

Even I don't like such questions .But these type of problems occur quite often in the exam.So I need to prepare for them.

Thanks for the input .

14. Mar 17, 2014

### haruspex

What can make these idealised questions hard is that by ignoring practical limitations the result can defy commonsense. In this question, the weightless spring has that affect. A real spring would 'use' some of its tension in its own acceleration.

Last edited by a moderator: Sep 25, 2014
15. Mar 17, 2014

### Rellek

Hey!

So, I really like this problem, but I've come across something interesting. If you add up the total length of the rope in the beginning you get:

L = l + x + S1 + S2

Where L = total length, l = the length of the spring if it is unstretched, and the other two portions of rope (not including the part that stays constant above and below the pulleys) denoted by S1 and S2.

So, I took a time derivative of this.

dL/dt = dx/dt + V1 + V2, keeping in mind l is constant.

But then I realized, the only portion of rope that is changing length is the portion of the spring that is being stretched. This means that dL/dt = dx/dt.

Does this mean that V1 = -V2, that simple? And from there a1 = -a2? What am I neglecting?

16. Mar 17, 2014

### haruspex

That there's a portion of rope vertically below the spring. This may start as zero but it won't stay zero.