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Homework Help: Spring pulley system

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data

    We have a system, consisting of a Disk radius R, with weight M, with a cordless string attached on its edge, at one edge of this cordless string we have a block of mass m, at the other end, we have the string attached to a spring with one end fixed to the floor, with spring constant k, also, the disk is on an axle, so it's mid air, and their is no friction between the disk and the axle. What is the max velocity of the block?

    2. Relevant equations

    Let's say the positive y direction is downward, we have the string to spring on the left, and the block of mass m attached to the right which will be released (at which the spring will also be in the relaxed state) so we choose the direction of the angular velocity to be clockwise, (towards suspended block)

    Fnet y of spring: T1 = kx

    Fnet y of Disk: T1 + T2 + Mg - N (axle on disk) = 0

    Fnet y of block mass M: mg - T2 = may

    tau (net) = T2R - T1R = Ialpha

    Vy^2 = Voy^2 + 2ay (delta y)

    3. The attempt at a solution

    We can find the acceleration of the block, by relating it with the following atan = ay = alphaR

    Thus alpha = ay/R

    Substituting alpha in Tau net we get

    T2R - T1R = MR^2(ay/R) ==> ( I used I = MR^2) , cancellation of R and solving for T2 = May + kx (I put T1 = kx)

    Plugged this in for Fnet of block mg - (May + kx) = may

    My result for ay = (mg - kx)/ (m + M) , now I can use Vy^2 to solve for a V but I get an equation in terms of X, m, M, g, K, and H (delta y) how do i go about finding Vy max of block with Voy = 0, at what point is the force of spring and force of w of block such that the block has a max v?

    My result ( I think it's wrong or I'm missing something) Vy = sqrt( 2ayh) using ay derived above..

    What am I missing to find the max Vy of block? What roles do mg and kx play in control of blocks speed, angular speed, and distance H displaced by block, also how can I relate delta x of spring (sorry I just realized its a bad idea to use x in kx, since we are only dealing with y dir. My apologies.) with the change in H of block? Can you? Thanks I advanced!
  2. jcsd
  3. Nov 27, 2014 #2


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    I can't decipher the description of the set-up. What's a 'cordless string'? In what way is it attached to the disk if it's also attached (at one edge?!) to a block and at one end to a spring? Is the disk vertical or horizontal? Do you mean the string runs over the disk, as in a pulley? Is the string vertical each side of the disk?
  4. Nov 27, 2014 #3


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    Is this the set-up?
    I think some data are missing. Is the spring relaxed when the block is released?
    When using conservation of energy, you have to include also the elastic energy of the spring. The formula Vy^2 = Voy^2 + 2ay (delta y) is not valid, as the acceleration is not constant.

    And welcome to PF :)
    Last edited: Nov 27, 2014
  5. Nov 27, 2014 #4
    My interpretation of the setup is the same as ehild's.
    I agree with your acceleration (though I would have used ##I = \frac{1}{2}MR^2## for the moment of inertia of a disk). If Vy is a maximum what does that say about ay at that instant? Once you apply that condition you can get the x value at which Vy_max occurs. If you have a relation between Vy and x then you can plug in your x value and find Vy_max. But as mentioned above, you can't use the constant acceleration kinematic equations.

    One way to get a relation between Vy and x is to use energy conservation between the initial state and Vy_max case. I think you'll have to make some assumption on the initial x of the spring (probably that it is relaxed and xo = 0). Don't forget that x and H are related to each other, so that's another variable you can substitute out.
  6. Nov 27, 2014 #5
    My apologies harupex, i was writing this on my ipad at the time, i meant massless, not cordless, so that T1s aren't affected by the mass of the string, Ehild is the interpretation is the one i'm talking about. and yes, the condition is such that the spring is relaxed when the block is released.

    Thank you :) it's a pleasure.

    Thank you folks, ill rework the situation and get back to you asap, happy thanksgiving!
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