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Work due to tension 
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#1
Nov406, 10:24 AM

P: 18

Hi,
When you hang a mass on a pulley and allow the mass to fall, do you normally have to take into account the work due to tension in the equation U1 + K1  W(tension) = U2 + K2??? I laways thought you did, but it doesn't take W(tension) into account in one of my problems. Thanks so much. Ryan 


#2
Nov406, 01:24 PM

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#3
Nov406, 02:28 PM

P: 18

Hmm, that is definitely not what they do here. Are they just purposely ignoring it? Here is the problem:
Two metal disks, one with radius R1 and mass M1 and the other with radius R2 and mass = M2 , are welded together and mounted on a frictionless axis through their common center. (see attachment) I=moment of intertia A light string is wrapped around the edge of the smaller disk, and a block with mass = m, suspended from the free end of the string. If the block is released from rest at a distance h above the floor, what is its speed just before it strikes the floor? The answer they got used the following formula: K1(=0) + U1 = K2,cylinder + K2,mass + U2 (=0) I don't see work due to tension anywhere. Is there something I am missing? Thanks. Ryan 


#4
Nov406, 05:21 PM

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Work due to tension



#5
Nov406, 07:35 PM

P: 312

In this case, conservation of energy is used as an alternative to Newton's second law. Depending on what you know, either will work. The only thing you need to solve for when using energy is angular velocity, which you can put in terms of translational velocity of the block



#6
Nov406, 09:43 PM

P: 18

So what you're saying is that usu., we don't need to take work by tension into consideration when dealing with problems where a mass is lowered from a pulley system? I knew we could ignore work due to tension if we had a pulley with a mass on each side, since the tensions cancel each other out, but I was never aware that the tensions cancel each other out when you have a pulley with a single mass hanging from it. So am I correct in my understanding: what you are saying is that in the situation where a single mass is lowered from a pulley, the work due to tension still cancels out since there is positive work in the direction of the dropping mass and negative work in the direction of the pulley? Thanks.
Ryan 


#7
Nov506, 11:57 AM

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If the string is stretchy, but perfectly elastic (no energy lost to heat) the motion could be far more complicated, with some of the energy being stored in the stretched spring. Then the work done on the block and the pulleys for some period of time would not necesaarily cancel, but the difference would be the energy stored in the string. If the string had mass, but was not stretcy, the tension would not be the same on the two ends of the string, so again the work done on the block and the pulleys would not cancel, but in this case the difference would be the kinetic energy gained by the moving massive string. These last two paragraphs, and a combination of both mass and stretching are a complication you need not encounter when first dealing with these problems. That is why the problems usually say "light string" and they expect you to assume that a string does not stretch unless you are given information about how it stretches in response to an applied force. 


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