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Releasing a rod tied to a string which was held at an angle

  1. Feb 25, 2016 #1
    1. The problem statement, all variables and given/known data
    "Find the tension in the right string, when the left string is cut, if the mass of the rod is ##m## and its length is ##l##, and ##\sin \theta = 3/5##. The rod is initially horizontal." Please refer the diagram below (sorry for its clumsiness!).
    PicPhys.png

    2. Relevant equations
    $$I_{rod} = \frac{ml^2}{12}$$
    $$\tau _{CM} = I_{rod,CM} \cdot \alpha _{CM}$$ ##\tau## represents torque, CM to denote "in the centre of mass frame", ##\alpha## is the angular acceleration.

    The Newton's Laws of Motion.

    3. The attempt at a solution
    First, only looking at the translational motion of the centre of mass, we get two equations:
    $$ mg - T \sin \theta = ma_x \qquad\text{...(i)}$$
    where ##T## represents the tension in the string and ##a## represents acceleration.
    $$ T \cos \theta = ma_y \qquad \text{...(ii)}$$
    Now, using the rotational dynamics formula,
    $$ T \sin \theta \cdot l = \frac{ml^2}{12} \cdot \alpha \qquad \text{...(iii)} $$
    I have these three equations, which i think are correct. However i doubt my fourth equation:
    $$ a_y - \frac{l\alpha}{2} = a_x \cot \theta \qquad \text{...(iv)}$$
    as acceleration of the point attached to the string, along the string, must be zero. However, on solving them, i get,
    $$T=\frac{mg}{2 \sin \theta - 3 \sin \theta \tan \theta}=\frac{-20mg}{3}$$
    which seems to be wrong (i don't have the correct answer to it). Where am i going wrong? Is the fourth equation really correct? If not, how do i correct it? Please guide me.

    Thanks in advance!
     
  2. jcsd
  3. Feb 25, 2016 #2

    haruspex

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    Your eqn (iv) looks ok, but the use of x and y there (standard) conflicts with your use in earlier equations (non-standard).
    Also, the left hand side of (iii) is out by a factor.
     
  4. Feb 25, 2016 #3
    Ah, i get it now! The first equation should have had ##a_y## and the second, ##a_x## (i did it by mistake). And the 3rd equation should have been:
    $$ T \sin \theta \cdot \frac l2 = \frac{ml^2}{12} \cdot \alpha \qquad \text{...(iii)} $$
    Now the answer comes as:
    $$ T=\frac{mg}{4 \sin \theta + \cos \theta \cot \theta}=\frac{15mg}{52} $$
    Is it correct now?

    Thanks again for your help!
     
  5. Feb 26, 2016 #4

    haruspex

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    I get the same.
     
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