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Components of torque in a solid sphere

  1. Dec 16, 2013 #1
    A uniform sphere of mass M and radius R has a point on its surface fixed at the origin. Its centre lies along a line in the direction of the position vector r = i + 2k + 3k at length R. Find the components of the torque acting on it due to gravity if the z-direction is upwards and gravity acts downwards.

    Solution:

    First thing I noticed was that there was no y component in the vector r so I figured perhaps there is no torque in that component!

    I tried τ= r x sinθ

    And simply used r as 5k for the z component and i for the x component!

    I used mg as the force and used θ as 90 because all axis are mutually perpendicular, but I think this is the wrong approach! I think the vector r would play more of role!

    Also would I use τ= Iαθ

    Because it's a sphere? We only touched on rotational torque very briefly in lectures and for this question there is no model answer!:(


    Thanks for any help:)
     
  2. jcsd
  3. Dec 16, 2013 #2
    It seems there is a typo in the position vector 'r' given in the question.It should be r = i + 2j + 3k .

    What is the position vector of the point of application of gravity ?
    Write the force of gravity in vector form .
    Calculate the cross product of 'r' and 'F' .This would give you the torque .
     
  4. Dec 16, 2013 #3
    The vector for gravity would be

    mg in the downward k direction

    So -2k?
     
  5. Dec 16, 2013 #4
    Why -2k ?
     
  6. Dec 16, 2013 #5
    If the torque due to gravity is z downwards! Is the torque in the positive z direction 0 as they are parallel or antiparallel rather
     
  7. Dec 16, 2013 #6
    You are misinterpreting the question.

    The question asks you to consider positive direction of z-axis upwards and gravity downwards . If the positive z-direction is ## \hat {k} ## ,then gravity points in ## -\hat {k} ## .

    What is the magnitude of force of gravity on the sphere ?
    How do you represent force of gravity in vector form ?
     
  8. Dec 16, 2013 #7
    The force =mg -z?
     
  9. Dec 16, 2013 #8
    No...

    ## F = -Mg\hat{k} ## .
     
  10. Dec 16, 2013 #9
    Sorry I didn't mean z!!
    Do I use τ = r x Fsinθ
     
  11. Dec 16, 2013 #10
    You could but will be a roundabout way. You don't know the angle so you will need to find the angle first.
    It would be easier to just cross multiply the two vectors in their component form.
     
  12. Dec 16, 2013 #11
    Do the cross product of the two vectors?
     
  13. Dec 16, 2013 #12
    And my component vector for force would just be -mg k?

    So there would be no torque in the x and y
     
  14. Dec 16, 2013 #13
    It will be. It's a cross product.
    What is i x j, for example? Is not the same as i.j (dot product).
     
  15. Dec 16, 2013 #14

    The way I was taught to do the cross product was in the form of a matrix, not sure!

    i x j is equal to z isn't it?
     
  16. Dec 16, 2013 #15
    i x j = k
    j x k =i
    k x i = j

    And if the order of the terms is changed, you have a minus sign.
    For example, j x i = -k.
     
  17. Dec 16, 2013 #16
    So for the x component I would do

    τ= r x F

    And use just the i component of the vector r?
     
  18. Dec 16, 2013 #17
    Why just the i component?
    Multiply term by term.
    Like you do in algebra when you have something like (a + b +c)x(p+q+r).
    Then you will see what components are present in the result.
     
    Last edited: Dec 16, 2013
  19. Dec 16, 2013 #18
    But then the result will be the same for every component will it not? What is changing?
     
  20. Dec 16, 2013 #19
    I don't understand your question.
    Just do the multiplication.

    ( i + 2j + 3k)x(-mgk)

    and see what you get.
    You are worrying too much for a trivial question.
     
    Last edited: Dec 16, 2013
  21. Dec 16, 2013 #20
    I would get

    mg j - 2mg i + 0
     
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