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!

Rolling Ball and angle of incline

  1. Dec 22, 2013 #1
    1. The problem statement, all variables and given/known data

    A uniform solid sphere rolls down an incline without slipping. If the
    linear acceleration of the centre of mass of the sphere is 0.2g, then what
    is the angle the incline makes with the horizontal? Repeat for a thin
    spherical shell.

    2. Relevant equations
    [tex]\sum \tau = I \alpha[/tex]
    [tex]I_{ball} = \frac{2mr^2}{5}[/tex]


    3. The attempt at a solution
    Moments about the tipping point:
    [tex]mgrsin\theta = \frac{2mr^2}{5}\alpha[/tex]
    Don't think this is going anywhere but I can't think of any other way to include moments of inertia.

    Just thought, this moment of inertia is from the centre of the ball, not the tipping point. Correct?

    So, by the parallel axis theorem,
    [tex]mgrsin\theta = (\frac{2mr^2}{5} + mr^2)\alpha = \frac{7mr^2}{5}\alpha [/tex]

    Which hardly helps
     
    Last edited: Dec 22, 2013
  2. jcsd
  3. Dec 22, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    So far, so good. Now relate the linear acceleration of the center of mass to the angular acceleration.
     
  4. Dec 23, 2013 #3
    Like this? (found from one of your posts)
    [tex]a_t= \alpha r[/tex]
     
    Last edited: Dec 23, 2013
  5. Dec 23, 2013 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, exactly.
     
  6. Dec 23, 2013 #5
    I've just never seen that before.

    [tex]mgrsin\theta = \frac{7mr^2}{5} \frac{g}{5}[/tex]
    [tex]sin\theta = \frac{7r}{25}[/tex]
    Does it want it in terms of r or have I gone wrong?
    [tex]\theta = sin^{-1}(\frac{7r}{25})[/tex]
     
  7. Dec 23, 2013 #6
    Recheck your calculations.You have missed an 'r' while making substitution. 'r' will cancel out .
     
  8. Dec 23, 2013 #7
    Ahhh, of course. Thank you!
    I take it the tangential acceleration is in the direction of the linear velocity? And is the radial acceleration just the centripetal acceleration?
     
  9. Dec 23, 2013 #8
    Tangential acceleration and radial acceleration of what ? In this problem we are dealing with linear and angular acceleration of the sphere.
     
  10. Dec 23, 2013 #9

    Doc Al

    User Avatar

    Staff: Mentor

    That's the condition for 'rolling without slipping'.
     
  11. Dec 23, 2013 #10

    Doc Al

    User Avatar

    Staff: Mentor

    The condition for rolling without slipping relates the linear acceleration of the center of mass to the angular acceleration:

    [tex]a = \alpha r[/tex]
     
  12. Dec 23, 2013 #11
    Does that work as an inequality? i.e. object won't slip providing as long as this holds true [tex]a \leq \alpha r[/tex]
     
  13. Dec 23, 2013 #12
    No...If [itex]a \neq \alpha r[/itex] ,then slipping occurs .For rolling without slipping [itex]a = \alpha r[/itex]
     
  14. Dec 23, 2013 #13

    Doc Al

    User Avatar

    Staff: Mentor

    No, it's an equality. If [itex]a \ne \alpha r[/itex], then there is slipping.
     
  15. Dec 23, 2013 #14
    How would it slip if the linear acceleration smaller than the RHS? I can picture the ball accelerating so quickly that it begins to slip, but not the other way around.
     
  16. Dec 23, 2013 #15

    Doc Al

    User Avatar

    Staff: Mentor

    Imagine the ball spinning, without enough traction.

    In practice, if you were to start the ball from rest at the top of the incline and there were insufficient friction to prevent slipping, you would have [itex]a \gt \alpha r[/itex].

    Nonetheless, if you have rolling without slipping, then [itex]a = \alpha r[/itex].
     
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: Rolling Ball and angle of incline
Loading...