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A question on rotation

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data
    A solid uniform sphere of radius R and mass M rolls without slipping with angular velocity (w) when it
    encounters a step of height 0.4R . find the angular velocity immediately after inelastic impact with the rough step.



    2. Relevant equations



    3. The attempt at a solution
    By conserving angular momentum about the contact point with the rough step -
    MwR(R-h) + Iw = Iw'

    But the answer is coming wrong . Its answer is 5w/7
     

    Attached Files:

  2. jcsd
  3. Mar 20, 2012 #2

    tiny-tim

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    welcome to pf!

    hi nik jain! welcome to pf! :smile:
    yes, i make it 5ω/7

    (you did notice it's a sphere, not a disc?)

    show us your full calculations, and then we'll see what went wrong! :smile:
     
  4. Mar 20, 2012 #3
    MwR(R-h) + Iw = Iw'
    Mw0.6(R^2) + 0.4Mw(R^2) = 0.4M(R^2)w'
    w'= 5w/2

    Please tell the mistake
     
  5. Mar 20, 2012 #4

    tiny-tim

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    ah!
    wrong moment of inertia :redface:

    you need I about the centre of rotation :wink:
     
  6. Mar 20, 2012 #5
    Need a little bit more help

    what is the value of moment of inertia and it is about which axis ?
     
  7. Mar 20, 2012 #6

    tiny-tim

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    hi nik jain! :smile:
    you need the angular momentum, L, about the corner of the step …

    in other words, about a point (O) on the surface of the sphere

    since this isn't the centre of mass (C), there are two ways to calculate it …

    i] (this works for any point) L = ICω + m*OC x vc.o.m

    ii] (this works only for the centre of rotation) L = IOω

    (they work out the same because IO = IC + mOC2 … the parallel axis theorem)​
     
  8. Mar 20, 2012 #7
    http://www.infoocean.info/avatar1.jpg [Broken]welcome to pf!
     
    Last edited by a moderator: May 5, 2017
  9. Mar 21, 2012 #8
    Thanks a lot
    I get it .
     
  10. Mar 21, 2012 #9
    I have a question, not very familiar with rotational problems. Angular momentum before impact is angular momentum of the rotation plus the angular momentum from moving of the center of mass. How does the formula MwR(R-h) give it to me?

    Don't I need to consider the velocity perpendicular to the line connecting the point on the edge with the center of mass of the object and do that using trigonometry?
     
  11. Mar 21, 2012 #10

    tiny-tim

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    hi chingel! :smile:
    Mωr is the momentum (Mvc.o.m), and R-h is the perpendicular distance from the point to the line of the velocity of the c.o.m.
    do you mean the component of velocity perpendicular to that line? :confused:

    no, you need r x mv, = mvrsinθ

    your formula (i think) is mvrcosθ
     
  12. Mar 21, 2012 #11
    I'm having trouble understanding what does the MwR(R-h) give? Is it r x mv and how does it give that?
     
  13. Mar 21, 2012 #12

    tiny-tim

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    yes, r x mv

    ωR is the initial speed, v

    so MωR is the initial momentum, mv

    then you need the distance, r, by which the initial momentum "misses" the pivot point (the corner of the step) …

    the initial momentum goes through the centre of mass, and it's horizontal, so it misses the step by a vertical distance, r = R-h :wink:
     
  14. Mar 21, 2012 #13
    Ah, I see. The angular momentum is considered at the time when the center of mass would be just over the contact point with the step. Even though this is not possible, the angular momentum stays the same no matter during which time we consider it, since there is no torque. At that point, the velocity and radius are perpendicular, enabling an easy calculation. Do I understand it correctly?
     
  15. Mar 21, 2012 #14

    tiny-tim

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    i think so

    if you use the perpendicular distance, the calculation is easy, because the sin is 1

    if you use the actual distance to the centre of mass (or any other point on the line), you have to multiply by the sin of the angle
     
  16. Mar 21, 2012 #15
    I understand, thank you for the replies.
     
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