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The solution of a problem regarding center of mass

  1. Dec 24, 2017 #1
    • This post lacks the homework template because it was originally posted in a non-homework forum.
    A rod of mass M (=4m) and length L is placed over a smooth horizontal table. One end of rod is attached with a block of mass m through a light string as shown in the figure. The pulley is friction less and light. the acceleration of com of rod at the given instant is (g = acceleration due to gravity)

    a) g/2
    b) g/8
    c) g/5
    d) 2g/5

    that does not make sense, because i think the force on com is net force on object = mg = Ma (force of com by newton's law)
    (M = mass of com = 4m)
    therefore, a = mg/M = mg/4m = g/4 which is none of the above

    note:
    1. Ma = mg is true even if the body rotates
    2. It's reasonable to have no friction, since it is smooth
    3. The rod does not seem to have any of it's points fixed to table to revolve around of
    4. There was no use of the fact that force was perpendicular to rod as in the figure
     

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  3. Dec 24, 2017 #2

    Orodruin

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    What makes you think the force is equal to mg?
     
  4. Dec 24, 2017 #3
    yeah, you are right, that must have been tension... but how do you find the tension in the rope
    If i were to hold the string wouldn't i have to exert mg to hold it in place... thus the force exerted on me which is it's tension is mg.... why would that be different in this case where there is a rod instead
     
  5. Dec 24, 2017 #4

    Orodruin

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    You need to somehow relate the motion of the rod (both linear and rotational acceleration) to that of the hanging mass.

    The easy part is: how does the tension in the rope relate to the acceleration of the hanging mass? See if you can figure out what other questions are helpfulto ask.
     
  6. Dec 24, 2017 #5
    acceleration of hanging body is a = (g - T/m)
    that's what i make of one part of tension...
    concerning the rod....
    A(com) = T/4m
    [​IMG](about com) = (L/2)*(T) = Iα
    therefore α = (LT/2)/(ML^2/12) = 3T/2mL

    woa... does the solution have anything to do with the fact that length of string does not change?
     
  7. Dec 24, 2017 #6

    Orodruin

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    So how can you use these relations to find the tension (and hence the CoM acceleration)?

    Hint: You need one more relation that relates the CoM acceleration, the angular acceleration, and the acceleration of the hanging mass as you currently have 3 equations (a = g - T/m, A = T/4m, and α = 3T/2mL) and 4 unknowns (a, T, A, and α).
     
  8. Dec 24, 2017 #7
    Does it require the use of energy conservation?
    or does it come from the fact that the string length is constant?
    It doesnt seem to come from energy because it will introduce height, a new variable, right?
     
  9. Dec 24, 2017 #8

    Orodruin

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    What do you get if you use that the string length is constant?
     
  10. Dec 24, 2017 #9
    but that makes acceleration of hanging mass and tip of the rod the same? so that they make the rope constant in length...
    then i get a = Lα /2
    Is that the fourth equation i needed?
     
  11. Dec 24, 2017 #10

    Orodruin

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    No, the acceleration of the rod CoM does not need to equal the acceleration of the hanging mass as the rod can turn. The acceleration of the tip of the rod however ...
     
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