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!

Ball Falling

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid spherical ball is placed carefully on the edge of a table in the position shown in the figure. The coefficient of static friction between the ball and the edge of the table is 0.5 . It is then given a very slight push. It begins to fall off the table.

    Find the angle (in degrees)(with vertical) turned by the ball before it slips.
    2e19e72524.fbc259d1f7.UhpXQY.png


    2. Relevant equations


    3. The attempt at question

    I have came up with some equations. Let ##\theta## be the angle(with vertical) at which the sphere begins to slip.

    p3.png
    By conservation of energy
    ##mgr-mgrcos\theta =\frac { 1 }{ 2 } { I }_{ 0 }{ \omega }^{ 2 }##

    As ##v=r \omega##

    So ##g(1-cos\theta )=\frac { 7 }{ 10 } \frac { { v }^{ 2 } }{ r } ##

    By finding torque about the contact point

    ##mgsin\theta r=\frac { 7 }{ 10 } m{ r }^{ 2 }\alpha ##

    As ##a=r \alpha##

    So ##\frac { 5gsin\theta }{ 7 } =a##

    Also from FBD of the block

    ##mgcos\theta -N=m\frac { { v }^{ 2 } }{ r } ##

    and ##\\ mgsin\theta -\mu N=ma##

    I don't know if these equations are right. Please help.
     
    Last edited: Nov 9, 2014
  2. jcsd
  3. Nov 9, 2014 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I doubt that's right for the centripetal force on a sphere.
     
  4. Nov 9, 2014 #3
    Could please explain, why? How to proceed?
     
  5. Nov 9, 2014 #4
    Can this be solved by just considering the forces at the point of contact tangent to the sphere? That would give:
    sin theta = 1/2 cos theta at the time of slipping.
     
  6. Nov 9, 2014 #5
    Could you please show how did you find that.
     
  7. Nov 9, 2014 #6
    The frictional force at the point of contact would be mg cos theta x 1/2 at a tangent to the sphere and the gravitational force would be mg sine theta.
     
  8. Nov 9, 2014 #7
    If you can find the normal force at the contact, then it is simple. Your result follows from equating the normal force with the radial component of the weight, but how do you justify this?
     
  9. Nov 9, 2014 #8
    voko ,what do you think about my equations?
     
  10. Nov 9, 2014 #9
    I agree with haruspex.
     
  11. Nov 9, 2014 #10

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Not in the way you write, as the ball is not moving in a straight line.

    See haruspex.
     
  12. Nov 9, 2014 #11
    Could you please explain why that is not correct?:p
     
  13. Nov 9, 2014 #12
    Because that assumes that the entire mass of the ball is at its center.
     
  14. Nov 9, 2014 #13
    Consider the element of the sphere just touching the corner, it is stationary.
     
  15. Nov 9, 2014 #14
    Then how should I proceed?
     
  16. Nov 9, 2014 #15
    But the rest of the ball is not. How do you justify the insignificance of that?
     
  17. Nov 9, 2014 #16
    The sphere is rigid, you need to assume that the force on every single atom is equal.
     
  18. Nov 9, 2014 #17
    Why?
     
  19. Nov 9, 2014 #18
    Do you think that the force of gravity only acts on the centre of mass?
     
  20. Nov 9, 2014 #19
    Personally, I would use a co-rotating frame. In that frame, the ball is in equilibrium. The sum of real and fictitious forces is zero.
     
  21. Nov 9, 2014 #20
    Does that give a different answer?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Ball Falling
  1. Falling Ball (Replies: 6)

  2. A Falling Ball (Replies: 3)

Loading...