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Problem in a non-inertial frame

  1. Dec 13, 2006 #1
    hey guys!
    i have a problem in physics which i could'nt solve:

    a car is moving with constant linear acceleration a along horizontal x-axis. A solid sphere of mass M and radius R is found rolling without slipping on the horizontal floor of the car in the same direction as seen from an inertial frame outside the car.What is the acceleration of the sphere in the inertial frame?

    Could anyone help me with it please?

    thanks.
     
  2. jcsd
  3. Dec 13, 2006 #2
    How the sphere rolling?
    I mean its acceleration in car'frame.
    Generally speaking,its acceleration in the inertial frame =car'acceleration+sphere's acceleration in car'frame
    It's Gallilus theory
     
  4. Dec 13, 2006 #3
    the sphere is rolling without slipping....thats all that has been given in the problem....does a pseudo force come into play??
     
  5. Dec 13, 2006 #4

    Doc Al

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    No need for pseudoforces. Hint: Since it rolls without slipping, what must be the acceleration of the bottom surface of the sphere? Apply Newton's 2nd law for both translational and rotational motion.
     
  6. Dec 13, 2006 #5
    i tried. but i am not getting it.could u be more precise with it please.
    thanks.

    since it is rolling without slipping,static friction comes into play.
    let a(obs) be the acceleration observed in the inertial frame.let f be the force of friction.then
    f*r = 2/5 m*(r^2)*a(obs)*r

    now applying Newton's 2nd law
    m*(a - a(obs)) = f

    this step is a bit confusing and am getting the wrong answer.
     
    Last edited: Dec 13, 2006
  7. Dec 13, 2006 #6

    OlderDan

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    What Doc Al is suggesting is that you work entirely in the inertial observers frame of reference. The force of friction will accelerate the CM of the ball, and it will produce a torque that gives the ball angular acceleration. You need to find the relationship between the linear and angular accelerations based on the no-slipping condition.

    An analogous problem you may have seen is a disk with a string wound around it laying flat on a frictionless table. What happens if the string is pulled horizontally with a constant force?
     
  8. Dec 13, 2006 #7

    Doc Al

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    Good.
    Careful with this. The observed acceleration of the car is given as "a". Are you using "a(obs)" to be the acceleration of the sphere? (The acceleration of its center of mass, I presume?)

    Another place to be careful. What you need here is the sphere's angular acceleration about its center of mass.

    I guess I don't know what "a(obs)" means.

    The key, as OlderDan says, is this:
    Hint: Try to express the angular acceleration of the sphere in terms of the accelerations of the sphere and the truck.
     
  9. Dec 13, 2006 #8
    the a(obs) here is the required acceleration of the sphere with respect to the inertial frame.
    are the two relations i have mentioned then correct?
    please help with the second one.with that clear the problem will be simple to solve.
     
  10. Dec 13, 2006 #9

    Doc Al

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    If that's the case, then neither is correct.
    OK.
    This is Newton's 2nd law for translation. You have called the force "f" and the acceleration "a(obs)". Rewrite it.

    (But you need to redo your first equation also.)
     
  11. Dec 13, 2006 #10
    Can we use [itex]f_r=\mu mg[/itex] and substitute that in the equation
    [tex]f_r=\frac{I\alpha}{r}[/tex] and then use

    [tex]a_o-r\alpha=a[/tex] where [tex]a_o[/tex] is the acceleration of the CM of the sphere?
     
  12. Dec 13, 2006 #11
    thanks buddies.the answer correctly comes out to be a(obs) = (2/7)*a.

    the only problem was due to the relative acceleration between the sphere and the car.
    also friction need not be equal to (mu)*m*g.the force of friction gets canceled out.

    thanks anyway.
     
  13. Dec 13, 2006 #12

    Doc Al

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    Good!

    Right; it's needed to get the proper angular acceleration of the sphere about its center.
    Exactly right.
     
  14. Dec 14, 2006 #13
    :eek:

    I just computed it.But come out to be (2/5)*R^2*a.And I suppose the correct answer should include R.Is there something wrong with the correct answer?
    my course:
    friction force=Ia=(2/5)*m*R*R*a
    inertial force=friction force(It maybe relative with a course called theory mechanics )
     
  15. Dec 14, 2006 #14

    Doc Al

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    Your answer is not even dimensionally correct. The correct answer is independent of R.

    Try using: Torque = I alpha
     
  16. Dec 14, 2006 #15
    Indeed!
    But apply this the answer is 2/5*a?:confused: :frown:
     
  17. Dec 14, 2006 #16
    "Try using: Torque = I alpha"
    Indeed!
    By apply this the answer is 2/5*a?
     
  18. Dec 14, 2006 #17

    Doc Al

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    What are you using for alpha? Express it in terms of "a" and "a(obs)". Hint: It's not a/r. Alpha is the angular acceleration of the sphere about its center of mass.
     
  19. Dec 15, 2006 #18
    My course:
    friction force=I*alpla/R=I*a/(R^2)=(2/5)*m*a
    And I supposed friction force =inertial force.
     
  20. Dec 15, 2006 #19

    Doc Al

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    As I said earlier, alpha does not equal a/r. To find alpha, you need the acceleration of the sphere's surface with respect to its center.
    For some reason, you seem to want to view this from a non-inertial frame, which is OK. But if the inertial force equaled friction (the only two forces on the sphere), the sphere would be in translational equilibrium.
     
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