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Pulley and two masses

  1. Sep 7, 2017 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations


    3. The attempt at a solution

    When the right mass moves horizontally towards right , the verticle force acting on it is Mg - Tcosϑ , where ϑ is the angle which right string makes with verticle . The left mass experiences a force Mg - T . Since verticle force experienced by right mass is more , it moves closer to ground .Hence B) should be correct .

    Is that right ?
     

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  3. Sep 7, 2017 #2

    BvU

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    What do your relevant equations have to say about that ?
     
  4. Sep 7, 2017 #3
    Motion of masses in vertical direction are governed by respective net forces acting on them in vertical direction .The corresponding net forces have been mentioned in the post .

    Is there anything more you would like me to explain ?
     
  5. Sep 7, 2017 #4

    BvU

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    Yes. I don'see the speed that is imparted to the mass appearing in your reasoning. So if the speed is zero, does it stilll apply ?
    And you don't say what T is either ....

    Did you try to do the experiment ?
     
  6. Sep 7, 2017 #5
    Tension in the string connecting the two masses .
    Why would speed appear in force equations ?

    We may consider any arbitrary horizontal speed being imparted to the right mass .
     
  7. Sep 7, 2017 #6
    Please write down your force balance in the horizontal direction for the mass on the right.
     
  8. Sep 7, 2017 #7
    Tsinθ = Max

    T is tension in the string and θ is angle which right string makes with vertical .
     
  9. Sep 7, 2017 #8
    At the instant after the mass on the right has received its horizontal velocity, what is the angle theta, and what is its centripetal acceleration with respect to the point of contact of the string with the pulley?
     
  10. Sep 7, 2017 #9
    zero .
    v2/L

    L is half length of string .
     
  11. Sep 7, 2017 #10
    OK. So, initially the vertical force balance on the right mass should really be: $$T-mg=m\frac{V_h^2}{L}+m\frac{dV_v}{dt}$$where ##V_h## is the initial horizontal velocity of the right mass and ##\frac{dV_v}{dt}## is the rate of change of its vertical velocity component.
     
  12. Sep 7, 2017 #11
    Ok . What should be the next step in the analysis ?
     
  13. Sep 7, 2017 #12
    Write the force balance on the left mass.
     
  14. Sep 7, 2017 #13
    T - Mg = May
     
  15. Sep 7, 2017 #14
    So, what do you get if you subtract the two equations, recognizing that, for the left mass, ##a_y## is the same as ##-dV_v/dt## for the right mass?
     
  16. Sep 7, 2017 #15
    ##\frac{dV_v}{dt} = - \frac{V_h^2}{2L}##
     
  17. Sep 7, 2017 #16
    Good. So what does this tell you about the initial vertical movement of the right- and left masses?
     
    Last edited: Sep 7, 2017
  18. Sep 7, 2017 #17
    Lengthening and shortening of a string is given by the expression ##\frac{dV_v}{dt}## .Isn't it same for both the masses ?
     
  19. Sep 7, 2017 #18
    No. They are moving in opposite directions. The total length of string is constant, so the length of string on the right gets longer, and the length of string on the left gets shorter.
     
  20. Sep 7, 2017 #19
    Can I ask you where did you take this problem?
     
  21. Sep 7, 2017 #20
    Since you took positive upwards and from the above expression of vertical acceleration of right mass has a negative sign , it's initial acceleration is downwards .Likewise acceleration of left mass is positive means it's acceleration is upwards .

    Right ?
     
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