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A problem from conservation of energy

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi everyone, my teacher gave this problem as an extra credit for last exam, which I just did very bad.

    The figure is attached to the post.
    There are 2 objects, which have the same mass (m). 2 objects are connected to each other by a string length (L), which is massless.
    At initial, the they are sitting on a vertical wall, no velocity, and there is NO friction.
    It start sliding down to another wall which is 30 degrees to the horizontal.

    So, we just have (m): mass of object 1 and 2, (L): length of string, and 30 degrees is the angle between horizontal plan to the wall it start sliding down.

    2. Relevant equations
    Determine the velocity of 2 objects at some time after the motion (there is no exact point of time, it could be "t")
    The hint is: object #v1 object #2 have the same velocity. (because they have a string between)

    3. The attempt at a solution
    I tried to do it by using conservation of energy, but I cannot go all the way because I can just figure out one equation but there is 2-3 unknowns.
    However, by doing other homework, I pretty sure that there is no normal force of the wall acting on 2 objects. (because if you do some home work, the result are usually v=sqrt(2gh). And my teacher also confirmed so.
    Please help me, i have 2 days left before it due.
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2014 #2

    ehild

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    Are you sure? That string can be slack. Which mass travels with bigger speed initially?

    ehild
     
  4. Mar 1, 2014 #3
    it is initially at rest. I dont really know why v1 and v2 is equal, but my prof said that v1 and v2 are equal. so it just 1 velocity for both of them.
     
  5. Mar 1, 2014 #4

    ehild

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    You can check with an experiment. You certainly have some toy cars. Connect two of them with a string and make the arrangement as in the figure - with a big book, for example, and see if the string stays slack or not, and the cars move with the same speed, when one moves still vertically downward and the other moves on the slope.



    ehild
     
    Last edited: Mar 1, 2014
  6. Mar 1, 2014 #5
    Oh, I forgot, that is a rigid string. moreover, my teacher just post the answer today, and he just ask how to derive to this equation: v1=v2= sqrt[gL (1-1/(2sqrt{3}))]
     
  7. Mar 1, 2014 #6

    ehild

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    So the two masses are connected with a stick instead of a string.

    Even then, neither the velocity, nor the speed of the two masses are equal in general. Think what are the velocities/speeds when the stick is rotated about one end.

    You gave the speed as a number, but the original problem asked the speed as function of the time.

    Have you copied the problem correctly?


    ehild
     
  8. Mar 2, 2014 #7
  9. Mar 3, 2014 #8

    ehild

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    In that problem, the position of the masses were P1(0,y1) of the mass moving on the vertical wall, and P2(x2,0) of the mass moving on the horizontal floor.

    The distance between them was equal to the length L of the stick: y12 + x22=L2.
    Taking the time-derivative,

    [itex]2 y_1 \dot y_1 + 2x_2 \dot x_2 =0 [/itex] *


    The time derivative of a coordinate is equal to the velocity component. Mass 1 moved vertically downward with velocity v1, [itex] \dot y_1 = v_1 [/itex], and mass 2 moved horizontally with velocity v2 to the right, , [itex] \dot x_2 = v_2 [/itex].

    The speeds were equal, so v1= - v and v2=v. Subbing into eq. *

    [itex]-y_1 v + x_2 v =0 [/itex] , that is, x2=y1. The rod made 45°angle with both the wall and the floor when the speeds of the masses were equal.

    According to the formula your professor gave, it is the speed when both masses move with the same speeds. That happens at a certain position of the rod.

    ehild
     
    Last edited: Mar 3, 2014
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