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!

Rotational Energy/Speed of a System?

  1. May 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid sphere of mass 6.0 kg is mounted on a vertical axis and can rotate freely without friction. A massless cord is wrapped around the middle of the sphere and passes over a 1.0 kg pulley and is attached to a block of mass 4.0kg. What is the speed of the block after it has fallen 80 cm? Treat the pulley as a sold cylinder.

    2. Relevant equations

    I of a solid cylinder = 1/2MR2
    I of a sphere = 2/5MR2
    KErotational = 1/2Iω2
    KElinear = 1/2mv2
    PE = mgh
    ω = v/r

    3. The attempt at a solution

    I am very lost in this question. I have tried to calculate the kinetic energies of each part of the system but it doesn't give radii or ω so I get stuck. I thought this could be relevant:

    KEsphere + KEpulley + KEblock

    But what would I equate those energies to? Thank you in advance for any help!
     

    Attached Files:

    Last edited: May 26, 2014
  2. jcsd
  3. May 26, 2014 #2
    I think I may have worked through it but it doesn't seem right:

    KEsphere = 1/2Iω2 = (1/2)(2/5mr22 = (1/5)(6kg)r2ω2
    then substitue in ω = v/r and get KEsphere =(1.2)vsphere2

    I did this for each the pulley and the block and got:

    KEpulley = (1/4)vpulley2
    KEblock = 2vblock2

    Then I thought I could use this equation:

    PEblock = KEsphere + KEpulley + KEblock

    Because the sphere and pulley don't have potential energy, right?

    mblockgh = (1.2)vsphere2 + (1/4)vpulley2 + 2vblock2

    (4kg)(9.81m/s2)(0.8m) = 1.2v2 + 0.25v2 + 2v2
    v = 3.0m/s

    The part I'm really not sure about is the velocities of each part of the system. Will they all be the same? I assumed they are all the same to get my answer
     
  4. May 26, 2014 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Of course, those are two different radii, and two different moments of inertia. -- Use subscripts or something similar.

    The image is below.

    I'll answer your 2nd post presently.

    attachment.php?attachmentid=70071&d=1401150325.png
     
  5. May 26, 2014 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The sphere and pulley are stationary; they don't have velocity.

    The only objects in this problem with linear velocity are the block and cord. The cord presumably doesn't slip on the surfaces of the pulley or the sphere.

    What do you think the velocities you used are relevant to?
     
  6. May 26, 2014 #5
    Are the velocities relevant to the speed the sphere turns at and the speed the pulley turns at?
     
  7. May 26, 2014 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Those velocities are the velocities of the surfaces of the speed & pulley at the points they contact the cord. Thus they're the velocity of the cord, which happens to be fastened to the block. Right >\?
     
  8. May 27, 2014 #7
    Oh ok. So because the cord is attached to the block, they should both have the same velocities?
     
  9. May 27, 2014 #8

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes.
     
  10. May 27, 2014 #9
    Thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted