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Rolling without slipping force diagram

  1. Nov 19, 2013 #1
    This is the second part of a 2 part problem. The first part is:
    A solid disc is rolling down a 30 degree incline from rest.
    a. draw the force diagram for the disk.
    b. write the equations for both translation and rotational motion
    c. Find the linear acceleration of the center of mass.
    d. If the radius of the disk is r, find the angular acceleration of the disk.
    e. If the mass of the disc is m, what is the expression for the friction force?

    I solved b. as: Translation {mgsin[itex]\varphi[/itex]-fs = ma , N=mgcos[itex]\varphi[/itex]} Rotation: fr=I[itex]\alpha[/itex]

    For c my answer was a=g/3

    For d I got [itex]\alpha[/itex]=2f/mr

    And for e I got f= mg/6

    Now part 2 says, "In the previous problem the disk started to roll from a height of h=2m with no initial velocity. Use the answers from part 1 to find:
    a. The time to reach the bottom of the incline.
    b. The velocity of its c.m. at the bottom of the incline.
    c. The angular speed of the disk at the bottom of the incline if the radius of the disc is 25cm.

    This is where I am stumped. I tried solving for the velocity by using Kinetic Energy, mgh = 1/2 I[itex]\omega[/itex]^2 + 1/2mv^2

    But I didn't get the correct answer. So I'm really lost, I tried Kinematic but just got lost not sure which equations to use.
     
  2. jcsd
  3. Nov 19, 2013 #2

    tiny-tim

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    Hi BrianSauce! Welcome to PF! :smile:

    (try using the X2 button just above the Reply box :wink:)
    That should work … show us your calculations. :smile:

    However, why didn't you just use the acceleration from part 1, as the question tells you? :confused:
     
  4. Nov 19, 2013 #3
    My calculations are mgh=1/2(1/2mr^2)(v^2/r^2) + 1/2mv^2
    gh=1/4v^2 + 1/2v^2
    gh=3/4v^2
    4gh/3 all under square root = v
    However, that was not the correct velocity. And I'm not sure what to do with the acceleration because to find time I need vf which I don't have.
     
  5. Nov 20, 2013 #4

    tiny-tim

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    HiBrianSauce! :smile:

    (just got up :zzz:)
    that's what i get also (with h = 2) :confused:
    but you can find the distance (along the slope), and use one of the other standard constant acceleration equations :wink:
     
  6. Nov 20, 2013 #5
    My lab partner had helped me, he used the kinematic equation Vf = 1/2at^2

    It completely slipped my mind. Thank you very much for the replies. :)
     
  7. Nov 21, 2013 #6

    tiny-tim

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    that doesn't look right :redface:

    there's s = vit + 1/2 at2 and vf2 = vi2 + 2as​
     
  8. Nov 21, 2013 #7
    Whoops, I meant delta x = 1/2at^2, no vi because it starts from rest.
     
    Last edited: Nov 21, 2013
  9. Nov 21, 2013 #8

    tiny-tim

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  10. Nov 21, 2013 #9
    Thanks again!
     
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