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Pulley and velocity problem

  1. Nov 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Two masses 53gms and 45gms are connected by a light inextensible string passing over a smooth pulley and hung freely. Find
    a) the velocity after 3 seconds
    b) distance traversed in 3 seconds
    c) distance traversed in the third second
    d)if the string is cut after 3 seconds how much farther will the masses move in the next one second
    2. Relevant equations
    common acceleration, a= (m1-m2)g/m1+m2
    S=ut+1at^2/2, v=u+at, v^2=u^2+2as

    3. The attempt at a solution
    I solved a, b and c but what I am finding difficult to solve is part d. The common acceleration gained after 3 seconds is 0.8m/s^2 and the velocity is 2.4m/s and the distance traversed in that time is 3.6m. When the string is in the normal state the end which is tied to the lighter mass will move up by 3.6m and the end tied to the heavier mass will go down by 3.6m. When the string is cut the initial velocity becomes 2.4 and final velocity becomes 0. The height through which the lighter mass moves is 0.298m. But the answer is 2.5m Tell me where i am going wrong..
     
  2. jcsd
  3. Nov 8, 2013 #2

    Doc Al

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    What makes you think the final velocity (after 1 second) is zero?

    When the string is cut what happens to the acceleration of the masses?
     
  4. Nov 8, 2013 #3
    The velocity will start decreasing once the string is cut until it finally gets to zero and there won't be any external force acting on the two masses apart from their own weights when the string is cut . So, the acceleration in this state is only g. What I am trying to figure out if the final velocity does reach zero in the next 1 second.
     
  5. Nov 8, 2013 #4

    Doc Al

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    One mass will slow down and the other will speed up, once the string is cut.

    Right!

    No reason to think so. But since you know the acceleration and the initial velocity, you can calculate the final velocity after 1 second.

    But you don't have to do that to answer part d, which asks for the change in position after 1 second.
     
  6. Nov 9, 2013 #5
    Well, the end with the lighter mass will slow down and the end with the heavier mass will speed up. Okay, Here is how I found out how much farther the heavier mass goes down. Since the heavier mass will speed up it's final velocity will be greater than it's initial velocity. So, v=2.4+9.8*1=12.2m/s Again (12.2)^2=(2.4)^2+2*9.8h then h=(12.2)^2-(2.4)^2/19.6=7.3m. Can I find out the change in position for the lighter mass using the same logic???
     
  7. Nov 9, 2013 #6

    haruspex

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    Yes, but what are your equations in this case? (Be careful with the signs.)
     
  8. Nov 9, 2013 #7

    Doc Al

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    OK. You can also find the change in position in a single step (using a different kinematic formula) without having to first figure out the final velocity. But your method is fine.

    Sure. But as haruspex advises, be careful with signs.

    I suggest, at least until you know the stuff cold, that you stick with a standard sign convention. That way you'll be ready for anything.
     
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