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Angular acceleration of a rod

  1. Aug 10, 2017 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=4882fd61b6534ddf43b92f7e90f1d4d6.png

    Note : In the above setup the string lengths are unequal and the left angle is 30° and right angle is 60° .


    2. Relevant equations


    3. The attempt at a solution

    Just after the string is cut , writing force eq. for rod in vertical direction .

    ##Mg - Tcos60° = Ma_y ## (1)

    Writing force eq in horizontal direction

    ##Tsin60° = Ma_x## (2)

    Writing torque eq about the CM of the rod ,

    ##Tcos 60°\frac{L}{2} = \frac{ML^2}{12} \alpha ## (3)


    Writing constraint eq. for the left end point of the rod where string is attached ,the component of acceleration of left end along the string length should be zero .

    ##a_ycos60° - \frac{αL}{2}cos60° - a_xcos30°= 0 ## (4)


    Solving the above four eqs .give ##T= \frac{2}{7}Mg##
    and ##\alpha = \frac{6g}{7L}##

    Could someone check my work ?
     

    Attached Files:

    • rod.png
      rod.png
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  2. jcsd
  3. Aug 10, 2017 #2

    haruspex

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    It all looks ok to me.
     
  4. Aug 10, 2017 #3
    According to official answer key , T=Mg/2 and α = 3g/2L :oldeyes: .

    Do you mind rechecking the constraint equation ?

    Is there some sign error ?
     
    Last edited: Aug 10, 2017
  5. Aug 10, 2017 #4
    Now consider a modified setup .
    ?temp_hash=3e915a2f9a266e4bf1f683613c191957.jpg
    The same four equations in this case would be

    ##Mg - Tsin2\theta = Ma_y##

    ##Tcos 2\theta = Ma_x ##

    ##Tsin \theta \frac{L}{2} = \frac{ML^2}{12}\alpha##

    ##a_y cos(90°-2\theta) - \frac{\alpha L}{2}cos(90°- \theta) - a_x cos2\theta = 0 ##

    Please check the above equations .
     

    Attached Files:

  6. Aug 10, 2017 #5
    I rechecked my calculations but the result is same as the one in OP .

    The answer key doesn't have too many mistakes .
     
  7. Aug 10, 2017 #6

    TSny

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    I also agree with the answer in the OP.

    One way to get their answer is to replace the correct constraint equation with the incorrect equation ##a_y = \alpha L/2## (as though the rod is instantaneously rotating about a fixed left end).
     
  8. Aug 10, 2017 #7
    Or in the case when both the strings are vertical .

    Do you think my equations for the set up in post 4 are correct ?
     
  9. Aug 10, 2017 #8

    TSny

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    Yes, a vertical string would give a constraint of ##a_y = \alpha L/2##. But then, I think you would get T = Mg/4.

    Yes, they look correct to me.
     
  10. Aug 10, 2017 #9
    Yes .

    Thank you very much .

    Thanks @haruspex
     
  11. Aug 15, 2017 #10
    Please see the attached image .

    Just after the right string is cut , the net acceleration of the CM of the rod is vertically downwards .Hence rod must move such that CM falls straight down.The left end must also fall straight down .

    But net acceleration of the left end point will be towards left ,which means the next instant after right string snaps , it moves towards left .

    Aren't the movement of CM and left end contradicting each other .

    I am sure I am messing somewhere .
     

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  12. Aug 15, 2017 #11

    haruspex

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    Why? There will be an angular acceleration.
     
  13. Aug 15, 2017 #12
    Right . So the next instant , rod moves such that CM moves straight down and the left end moves leftwards ?
     
  14. Aug 15, 2017 #13

    haruspex

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    Yes.
     
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