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Angular Momentum of mass

  1. Jul 4, 2015 #1
    1. The problem statement, all variables and given/known data
    screenshot_6.png

    2. Relevant equations
    L = mvr
    L = Iw

    3. The attempt at a solution
    I did not attend this lesson due to some reasons. I read it from the book but I could not understand it well. I know the linear momentum well. However, in angular momentum, we have two equations I don't know which one is for which.

    My Attempt:

    Before:
    We have only momntum for the mass that is moving..
    L = mvr = 1 × 1 × 0.5 = 0.5

    After
    L = (Iball + Irod) w = w(Iball + 1/12)

    That''s all what I could done
     
  2. jcsd
  3. Jul 4, 2015 #2

    Andrew Mason

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    You have the right idea:

    Lbefore = Lafter where L is measured relative to the axis of the rod.

    It is just a matter of setting up the right side.

    Lafter = Lball + Lrod

    Can you complete the right side? (hint: you will have an expression in which the only unknown is ##\omega## the angular speed of the ball and rod after collision).

    AM
     
  4. Jul 4, 2015 #3
    Thanks,
    Lrod = Irod * w = w * ML2/12 = w/12
    but what about the Lball .. what forumla should I use?? L = mvr or L = Iw (can you help me to distinguish from them)
    If L = Iw then what is I ? should it be mr2 ?
     
  5. Jul 4, 2015 #4
    Paying more attention to your units should help.

    It is good practice to never write numbers without their corresponding units when solving a physics problem.
     
  6. Jul 4, 2015 #5
    Thanks but I still don't know how to find angular momentum.
    Is it L = mvr or L = Iw
    which one to use and when
     
  7. Jul 4, 2015 #6
    OK, I solved it . The answer is 5.2
    But I don't know when to use L = mvr or L = Iw
    Thanks
     
  8. Jul 4, 2015 #7

    haruspex

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    The short answer is to use both. The motion of a rigid body can be expressed as the velocity of the mass centre plus a rotation about that centre.
    The linear part contributes the mvr term, while the rotation adds an Iw term.
    In many cases, one or other of these turns out to be zero.
    Alternatively, you may be able to express the motion of the body (instantaneously) as purely a rotation about some axis. E.g. for a rolling wheel, the instantaneous centre of rotation is the point of contact with the ground. In this case you may know the moment of inertia about that point. Having expressed the motion entirely as that rotation, the linear term vanishes.
     
  9. Jul 5, 2015 #8

    Andrew Mason

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    The equation ##L = I\omega## is always true for a rigid body. The equation ##L = mvr = mr^2\omega## is true only for point masses. For a point mass, L = mvr = ##m(\omega r) r = mr^2\omega = I\omega## So for a point mass rotating about an axis a distance r away, the moment of inertia is mr2. For calculus purposes, one can think of a rigid body as being made up of an infinite number of point masses distributed according to the geometry of the body.

    AM
     
    Last edited: Jul 5, 2015
  10. Jul 5, 2015 #9

    haruspex

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    That's over-restrictive. You can use mvr for the linear portion of the motion of a rigid body, as I described.
    Since it can be used for each mass element of the body, we can write ##L=\int vr.dm=v\int r.dm=vm\hat r##, where r for a mass element is the distance from that element's line of travel to the reference axis, and ##\hat r## is the distance from the mass centre's line of travel to the reference axis.

    Edit: and note that one should always think in terms of rotation about an axis, not rotation about a point. Otherwise you might compute the distance wrongly.
     
    Last edited: Jul 5, 2015
  11. Jul 5, 2015 #10

    Andrew Mason

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    If you use L = mvr for anything other than a point mass, how do you define r? If you use the distance of the centre of mass to the axis of rotation (of the rod), it is only approximate. In the case here, we are approximating the 50 g. mass as a point mass and use L = mvr. Although that is what the student is expected to assume here, it really isn't a point mass.

    If you want to be accurate you have to use the moment of inertia of the 50g mass, ie. L = ##I\omega## not L = mvr. This would involve determining the moment of inertia of the 50 g body about the axis through its centre of mass and parallel to the axis of rotation of the rod and then using the parallel axis theorem (to determine the moment of inertia about the axis of rotation of the rod). Obviously that is not what is expected here so we just approximate it using L = mvr and assume it is a point mass.

    Your note about rotation about an axis is well taken. I have edited my earlier post.

    AM
     
    Last edited: Jul 5, 2015
  12. Jul 5, 2015 #11

    haruspex

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    As I wrote, it should be the distance from the (instantaneous) line of travel of the mass centre to the reference axis (not necessarily the axis of rotation). This is not an approximation, as I thought I showed with my integrals.
    A useful application of this is for the standard question about a bowling ball which is initially sliding, not rotating. Taking moments about an axis on the ground, the initial angular momentum is mvr. When rolling, it is ##mv_fr+\frac 25 mr^2\frac {v_f}r=\frac 75 mv_fr##. Since friction has no moment about that axis, the initial and final are equal, yielding the final speed very neatly.
     
    Last edited: Jul 5, 2015
  13. Jul 6, 2015 #12

    Andrew Mason

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    The initial angular momentum is mvr. But if it collides with and sticks onto a rod and starts rotating about an axis through the middle of the rod (as in this problem), you cannot ignore how that the mass is distributed and just treat it as a point mass. In this case we approximate the mass as a point mass and ignore the way it is distributed.

    AM
     
  14. Jul 6, 2015 #13

    SammyS

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    What distribution are you referring to?
     
  15. Jul 6, 2015 #14

    haruspex

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    If a rigid body is rotating about a point other than its mass centre it is always valid to treat that as the sum of a rotation about its mass centre and the linear motion of its mass centre. You can apply mvr to the linear part, Iw to the rotational motion, and sum the two. This is entirely equivalent to using the parallel axis theorem.
     
  16. Jul 6, 2015 #15

    Andrew Mason

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    But in this case, after the collision there is no longer any linear motion. The body is stuck on the rod and is rotating about the axis through the geometric centre of the rod. In that case, you have to use the moment of inertia to find the angular speed: ##L_{initial} = m_{obj}vr = L_{final} = \omega({I_{obj}} + I_{rod}) ##

    My only point is that the moment of inertia of 50g object is not exactly mr2, although for this problem that is a good approximation. I depends on how that mass is distributed.

    AM
     
  17. Jul 6, 2015 #16

    haruspex

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    I feel you are not reading my posts carefully.
    The motion of the ball after sticking to the rod is a rotation about the rod's centre, not a rotation about its own centre.
    We can think of this as purely a rotation about the rod's centre, or we can think of its instantaneous motion as a linear (tangential) motion plus a rotation about the ball's centre. With the first option, we use the I given by the parallel axis theorem (I assume that is what you intended in your equation above); in the second, we use the I for rotation about the mass centre and add mvr.
    Since ##mv_{final}r=mr^2\omega##, the two methods are completely equivalent.
     
  18. Jul 7, 2015 #17

    Andrew Mason

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    Actually, it is a combination. The ball (ie. the 50 g. mass) is rotating about an axis through its own centre of mass at the same angular speed as its centre of mass is rotating about the axis of rotation of the rod.
    So it seems you agree that one needs to know the distribution of mass within the ball to make it exact (i.e. we have to know I for rotation about the mass centre). That was my point.

    AM
     
  19. Jul 7, 2015 #18

    haruspex

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    Of course I agree. It's consistent with what I have said in every post on this thread. Our dispute began (posts #9 and #10) when you objected to my using mvr for the linear portion of motion of a rigid body.
     
  20. Jul 7, 2015 #19

    Andrew Mason

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    I am referring to the dimensions and shape of the 50 g object that collides with the rod: for example, the mass might be a spherical ball of a some radius and of uniform density.

    AM
     
  21. Jul 7, 2015 #20

    Andrew Mason

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    ?? I was not objecting to using L = mvr for the angular momentum of a non-rotating rigid body moving in a straight line (such as the 50g object before it collides with the rod, relative to the axis through the centre of the rod). I was objecting to using L = mvr to determine the angular momentum of the body when it is rotating about any axis, such as the axis through the centre of the rod. I merely said that "The equation L=mvr (for a rigid body rotating about an axis such this 50 g mass rotating about an axis through the centre of the rod) .... is true only for point masses.". You said that statement was "over-restrictive"....

    AM
     
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