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Work done dynamics question

  1. Jan 1, 2015 #1
    Dear All, would really appreciate some help with the following question - am struggling with dynamics at the minute!
    1. The problem statement, all variables and given/known data

    An 8kg mass is suspended from a grooved drum at radius r1=200mm and with a speed of 0.3m/s it drops 1.5m.
    The outer radius of the drum is r2=300mm. Assume that the drum has radius of gyration k=210mm and mass 12kg and a constant frictional torque in its bearings of 3Nm. Determine the speed once it has dropped the 1.5m.

    2. Relevant equations
    So in a hint I have been told to utilise s=r x theta and also v=wr

    3. The attempt at a solution
    I attempted the following:
    mgh + 1/2m(mass)v^2 + 1/2 Iw(drum)^2 = 1/2m(mass)v^2 + 1/2 Iw(drum)^2

    With the left hand side being the initial velocities with the 0.3m/s.

    This threw me well out (the answer is supposed to calculate to 3.01m/s.

    Anyone able to help, it would greatly appreciated!

    Thanks!
     
  2. jcsd
  3. Jan 1, 2015 #2
    Sorry, forgot to add - I had a -3 in the left hand side as the friction.....
     
  4. Jan 1, 2015 #3

    BvU

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    Well, did you work out your attempt ? How did you wrangle your -3 into the left hand side ? Is the friction there only initially ?
     
  5. Jan 1, 2015 #4
    Hi BvU,
    I was using the theory that,

    T1 + ∑U1-2 = T2

    Hence the 3Nm frictional torque was placed as a minus from the mgh on the left hand side. The frictional torque is stated as constant in the question hence I understand that it has to included as part the energy equation.
    I did this problem earlier in the day and I belive I got something like 2 x10^-4 (!) so you can see I was well out....
    I'm going to try it again as possibly tonight or tomorrow morning as I am flats out with assignments and revision at the moment so will stick up the answer I get the 2nd time around - never kn ow, I might get it right!
     
  6. Jan 2, 2015 #5
    So to be a bit clearer on this and as sort of an answer to BvU, the following is the equation I have come up with to solve:
    https://pod51046.outlook.com/owa/service.svc/s/GetFileAttachment?id=AAMkAGY5NTlkMjQ3LTdkOTUtNGRhMS05YjFlLTVhNzkzMTliNzUzOQBGAAAAAABNCMhttiqcQLg6VZ%2FUAKBeBwCA0zEXai05TZTyoTdiil0LAAAAefBKAADPKniALmOrSrBCx4TcTWeaAAE1I%2B0BAAABEgAQALuE7DSqlMpKlAAKu8VK5ek%3D&X-OWA-CANARY=OHs_oKPiqEyBerIAxjKWY6DFHxa79NEIHWIw4x0FUc0LBuIuZ27LjwRQCZZZe69f5KNq6ky2dbw.
    However, I cannot solve as I cannot work out the linear acceleration to calculate M; (M = F*s and the F is calculated from the tension in the rope which is T=m(g-a).
    This as well as the unknown V^2 I (which is what I'm looking for) is really acting as a 'show stopper' for me.
    Anyone??
     
  7. Jan 2, 2015 #6
    Sorry, it wont let me post up the equation that I have copied and pasted from word!!
     

    Attached Files:

  8. Jan 2, 2015 #7

    Stephen Tashi

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    Why don't you type out a list of the numerical values you used in the equation?
     
  9. Jan 2, 2015 #8
    m(block)=8kg
    g=9.81m/s^2
    r1=0.2m
    r2=0.3m
    k=0.21m
    m(drum)-12kg
    Torque(tf)=3Nm
    u=0.3 m/s

    I=mk^2 was used to calculate I of the drum
    Theta was calculated from = s/r
     
  10. Jan 2, 2015 #9

    BvU

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    [edit] I started this yesterday but apparently forgot to post, sorry.

    Although it is tempting to consider the dimension of a torque to be equal to the dimension of an energy, this is not the case: a torque is a (pseudo) vector and an energy is a scalar. So you can't just subtract the 3 from Δmgh.

    Could you post your calculations in detail ? It enables more helpful assistance.

    [edit] Well, you did some of that, so that deserves a bit more help:

    I suppose the initial 1/2 mu2 gave you 0.36 J ? That's what I get.
    And the initial 1/2 I u2/r22 another 0.265 J ?

    As I said, the ##\tau_f## can't just be added to ##mgh##, which is a substantial 118 J . Strange you should get such low final speed !

    And what is ##M\theta## ? Same story as ##\tau_f## ? Didn't ##M\theta## change into ##{1\over 2} I\; \omega^2 \ ##?

    Now comes the bummer: (since you dropped the supposed answer) I can't reconstruct the 3 m/s from the energies (0.5 J + 118 J - Wf =? about 60 J from 3 m/s).

    Perhaps it's necessary to find the actual acceleration of the mass and use v = at with t from v0+at2 ?
     
  11. Jan 2, 2015 #10
    θHi BvU, thanks for the input.
    The question was handed in a tutorial on work done from uni with the answer given as 3.01m/s and I have since found this question in a Meriam textbook with the same answer but no worked example or one that is anything like it to draw from.

    I included the Mθ as I assumed this was a rotational energy, so yes , same as the torque - this is incorrect I take it?
     
  12. Jan 3, 2015 #11

    BvU

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    Rotational energy is already accounted for in ##{1\over 2} I\; \omega^2 ##, so this way you count it twice !

    Any ideas how to deal with the friction already ?

    And: -- just to be sure -- is the mass initially moving upward at 0.3 m/s or is it moving downward ?
     
    Last edited: Jan 3, 2015
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