1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring, Pulley Problem Help.

  1. Nov 1, 2012 #1

    MG5

    User Avatar

    A 3.8-kg sphere is suspended by a cord that passes over a 1.3-kg pulley of radius 3.5 cm. The cord is attached to a spring whose force constant is k = 86 N/m as in the figure below. Assume the pulley is a solid disk.

    8-p-089.gif

    (a) If the sphere is released from rest with the spring unstretched, what distance does the sphere fall through before stopping?

    The answer to this is .866 m. Not sure how to get it though.

    (b) Find the speed of the sphere after it has fallen 25 cm.

    Answer to this is 1.73 m/s. Not sure how to get this either.

    I'm not quite sure how to start this. I guess I'd use...

    1/2kx2 for the spring, mgh since the sphere has PE before falling, 1/2mv2 for final KE of the falling sphere. And I'd have to do something about the pulley. I guess KE=Iω2 since its rotating? Not sure about that. And its a sphere that is falling and the I for a sphere is I=2/5mr2. Dont know if thats needed for this problem. Any help would be great. Thanks.
     
  2. jcsd
  3. Nov 1, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi MG5! :smile:
    same method you're using for (b), with final speed 0 for both the sphere and the pulley :wink:
    1/22, otherwise yes :smile:
    no, you only need moment of inertia if the sphere is rotating :wink:
     
  4. Nov 1, 2012 #3

    MG5

    User Avatar

    Yeah I meant to add the 1/2 in there. Thanks for pointing that out. I'm gonna give it another shot and see how it goes.
     
  5. Nov 1, 2012 #4

    lewando

    User Avatar
    Gold Member

    For what its worth for (a), when the sphere loses the maximum PE, the spring will have gained maximum PE. Conveniently, at this time the pulley is not spinning. This does not help with (b) though.
     
  6. Nov 1, 2012 #5

    MG5

    User Avatar

    Ok this is the equation I came up with. Doesnt seem right though.

    mghi=1/2mv2+1/2kx2

    I dont know what the height would be though
     
  7. Nov 1, 2012 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    is this (a) or (b)? :confused:

    if it's (a), v = 0; if it's (b), what about ω ?
     
  8. Nov 1, 2012 #7

    lewando

    User Avatar
    Gold Member

    As long as the cord does not stretch x and h will be related.
     
  9. Nov 1, 2012 #8

    MG5

    User Avatar

    First I'm trying to figure out (a). Then I'll try to do (b)

    Well the 1/2mv2 was the final KE. If it was initial it would be zero. But yeah I just realized I dont have the velocity either. Too many unknowns
     
  10. Nov 1, 2012 #9

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    but the final velocity is also zero :wink:
     
  11. Nov 1, 2012 #10

    MG5

    User Avatar

    yeah I saw that but I still wasnt sure.

    So would it be something like mgh=1/2kx^2
     
  12. Nov 1, 2012 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    why??! :rolleyes:
    yup! :biggrin:
     
  13. Nov 1, 2012 #12

    MG5

    User Avatar

    I thought it wasnt just because in like projectile motion problems when it asks how far something traveled and like where it hit the ground from its initial starting point the final velocity was never zero. I thought we usually needed that value to find x. But thats a completely different problem. Now I know.

    Only problem with that equation is I have no idea what the height is to put into mgh.
     
  14. Nov 1, 2012 #13

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Set up an energy equation.
    Remember that mgh, FOR THE PULLEY,is a constant until "Dovre falls"
    (Dovre is a mountain chain in my native country Norway!)
     
  15. Nov 1, 2012 #14

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    same as x ! :smile:
     
  16. Nov 1, 2012 #15

    MG5

    User Avatar

    but then theres two unknowns, one on each side of the equation. Sorry if I'm just really off with this. Hopefully you can bear with me.
     
  17. Nov 1, 2012 #16

    lewando

    User Avatar
    Gold Member

    No worries! x = h. x can be substituted for h and vice-versa. [edit-- sorry]
     
    Last edited: Nov 1, 2012
  18. Nov 1, 2012 #17

    MG5

    User Avatar

    Yeah but still. It could be giraffe lol. Its still an unknown.
     
  19. Nov 1, 2012 #18

    lewando

    User Avatar
    Gold Member

    You have 2 equations and 2 unknowns.

    kx2/2= mgh

    x = h

    This is solvable.

    [edit-- I originally said x = y, I meant x = h]
     
    Last edited: Nov 1, 2012
  20. Nov 1, 2012 #19

    MG5

    User Avatar


    Yeah it is. Sorry I'm just not used to having to unknowns. As usual, I was making it much harder than it was.

    I got it

    3.8kg(9.8)x=1/2(86N/m)x^2

    37.24x=43x^3

    Subtract 37.24, then do 43/37.24, and you get the answer .866. That was easy.

    Now the second part.

    Finding the speed after it has fallen .25 meters.
     
  21. Nov 1, 2012 #20

    lewando

    User Avatar
    Gold Member

    What?
    There should be no subtraction and no x3.

    Unless that's a typo, you got lucky on the math. Make sure you are clear on this before proceeding.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Spring, Pulley Problem Help.
Loading...