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Direction of tension

  1. Mar 16, 2017 #1
    1. The problem statement, all variables and given/known data
    A block with mass m is revolving with linear speed v1 in a circle of radius r1 on a frictionless horizontal surface. The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to r2. (a) Calculate the tension T in the string as a function of r, the distance of the block from the hole. Your answer will be in terms of the initial velocity v1 and the radius r1. (b) Use W = ∫r1r2T(r) ⋅ dr to calculate the work done by T(r) when r changes from r1 to r2. (c) Compare the results of part (b) to the change in the kinetic energy of the block

    2. Relevant equations

    3. Attempt at solution
    I've solved the question except that I thought that T ⋅ dr = Tdr while it was supposed to T ⋅ dr = -Tdr because apparently T and dr are antiparallel. I thought that tension is only in the direction towards the middle of a rope and we're obviously pulling r closer to the middle so how come they are antiparallel?
     
    Last edited by a moderator: Mar 16, 2017
  2. jcsd
  3. Mar 16, 2017 #2

    Doc Al

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    Note that dr points in the direction of increasing r, which is outward.
     
  4. Mar 17, 2017 #3
    Oh. You mean dr is not just a change in length but specifically a positive change? So -dr would ALWAYS be a negative change?
     
  5. Mar 17, 2017 #4

    BvU

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    ##r_1## goes to ##r_2## for the integral. Since ##r_1<r_2##, that minus sign should come out of the integral.
     
  6. Mar 17, 2017 #5

    ehild

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    ##\vec r## is the position vector, pointing to the block, and ##\vec T ## is the force of tension which points towards the center. They are anti-parallel. ##\vec {dr} ## is the change of ##{\vec r }##. The work done by the tension is ##W =\int_{r_1}^{r_2}{\vec T \cdot \vec {dr}}## Written with the magnitudes T and dr, ##W = \int_{r_1}^{r_2}{T dr cos(\theta)} = \int_{r_1}^{r_2}{(-T dr) }##

    upload_2017-3-17_15-36-24.png
     
  7. Mar 17, 2017 #6

    BvU

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    Ehild is of course correct: ##\hat T## and ##\hat r## have to point in the same direction for the dot product to make sense.
     
  8. Mar 17, 2017 #7
    Thanks ehild. I fully get it now!
     
  9. Mar 17, 2017 #8

    kuruman

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    Another way to look at it

    For line integrals I find it least confusing to write the two vectors in the integrand in terms of unit vectors, then take the dot product, and finally do the integral. In this case, for example, I would write ##\vec{T} \cdot d\vec{r}=(-T)\hat{r} \cdot (dr)\hat{r} = -Tdr##. So then, ##\int_{r_1}^{r_2}{\vec T \cdot d\vec{r}} =- \int_{r_1}^{r_2}{T~dr}##. Note that ##d\vec{r}## is always ##+(dr) \hat{r}## regardless of whether the line integral is taken radially out or radially in. In other words, don't worry whether the two vectors are parallel or antiparallel; the minus sign, if needed, will be taken care of by the limits of integration.
     
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