# Work due to tension

1. Nov 4, 2006

### rleung3

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. Nov 4, 2006

### OlderDan

Yes you do. If it were not for the work due to tension, the mass would have the same acceleration as it would in free fall. However, most problems of this sort involve two masses, one on each side of the pulley with equal tension on both sides. The work done by tension on one mass cancles the work done by tension on the other mass. One is positive (tension in direction of motion) and one is negative (tension opposite direction of motion). So, if you calculate the total kinetic and potential energy for both masses, the work done by tension is not a factor.

3. Nov 4, 2006

### rleung3

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

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4. Nov 4, 2006

### OlderDan

I cannot yet see the diagram, but I get the idea. This is the same type of situation as with two masses on opposite sides of a pulley. The tension does positive work in one case (applying a torgue in the direction of rotation of the disks) and negative work in the other (applying a force oppostite the direction of motion of the mass). The work done by the tension cancels out. As long as you consider the total kinetic energy and potential energy of the system of mass plus disks, you do not need to calculate the work done by the tension.

Last edited: Nov 4, 2006
5. Nov 4, 2006

### turdferguson

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. Nov 4, 2006

### rleung3

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. Nov 5, 2006

### OlderDan

That is what I am saying. As long as the string does not stretch, the objects attached at both ends must move the same distance. Both objects experience the same magnitude of force, but one object is moving in the direction of the force and one is moving against the force. One object has work done on it, and one has work done against it. Work is just the mechanism by which energy is transferred from one object to another. In this problem, a lot of the potential energy that the block had is transferred by the string to the pulleys as kinetic energy of rotation, with some potential energy being converted to kinetic energy of the moving block. No energy is gained or lost by the string.

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.