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When to use parallel axis theorem for objects...

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  1. Dec 1, 2015 #1
    1. The problem statement, all variables and given/known data

    A uniform solid ball of mass m and radius R rolls without slipping down a plane inclined at an angle f above the horizontal. Find the frictional force and the acceleration of the center of mass.



    2. Relevant equations
    τ=I*α
    so:

    fs*r=I*a

    Mg-Fs=ma

    Moment of inertia for sphere about center of mass: 2/5*I

    Moment of inertia for rolling sphere: 7/5*I


    3. The attempt at a solution

    I understand how to do the problem, I am just confused about why exactly we use the center of mass moment of inertia for "I" (2/5*I) in newtons second law Torque equation. Is it just by definition...? the moment of a rolling sphere is actually 7/5*I according to the Parallel Axis Theorem. I understood it before I learned the Parallel theorem, now I am just a tad bit confused. An input would be great, thanks!!! :)

    *also, hopefully this is the correct format and everything, I am new to this site*
     
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  3. Dec 1, 2015 #2

    SteamKing

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    When the ball is rolling down the inclined plane, is it rotating about its center of mass, or is it rotating about some other point?
     
  4. Dec 1, 2015 #3
    Excellent, thanks for the response! My best guess is that the center of mass is only moving, position wise, in a straight line. It is moving along but not rotating around the incline...so the moment of inertia would be (2/5)*I. Thing is, I saw it calculated to be 7/5*I somewhere else! Just confused! :( Any additional input would be great :)
     
  5. Dec 1, 2015 #4

    SteamKing

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    You're still not understanding what I'm asking.

    When the sphere is rotating about an axis which passes thru its center, then the MOI is 2/5*MR2, where M is the mass and R is the radius of the sphere.

    When a sphere is rolling down an incline, it's not rotating about an axis passing thru its center, is it? Where does the rotation axis lie relative to the center of the sphere?

    Hint: It's OK to draw a picture.
     
  6. Dec 1, 2015 #5
    Yes, I understood you; I don't think you fully understand my question (which is at the end :) *Side note: I spent 15 minutes with a tutor today, and her explanation was inadequate, as she said that the "r" was zero, and therefore the moment would be just 2/5*I for a rolling sphere.
    Anyway, when I did what you said, the sphere appears to be rotating relative to a point on the bottom of the circle; it also rotates around itself. So, I would assume that the moment of the rolling sphere itself would be 7/5*I like I initially thought! This may indeed be correct, but therein lies the issue: the problem I was working on, using the formula for torque τ=ι*α, only wanted the moment of inertia for the center of mass, 2/5*I. Is that the standard convention for that formula? To only use the center of mass's inertia? I think this is the right answer. I really appreciate your response, StreamKing! :)
     
  7. Dec 1, 2015 #6

    SteamKing

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    The handle is SteamKing, not StreamKing, BTW.

    Your terminology is confusing. How can the MOI of a sphere be 2/5 I ? What is I? The MOI of a sphere rotating about its C.O.M. is (2/5)MR2.

    In any event, you should draw a free body diagram of the sphere on the incline.

    What is the frictional force between the incline and the sphere if the coefficient of friction is μ and the angle of the incline is θ ?
     
  8. Dec 1, 2015 #7
    haha sorry, I thought it was stream xD. I=m*r^2. I could have written it the other way (2/5*m*r^2), but I didn't because I assumed people would know what I was talking about (and it takes longer to write each time). Fs=Cos(θ)*μ. Anyway, yeah, I am a huge proponent of a heuristic approach, but your opinion would be great on the issue. And yes, I know the the MOI of a sphere rotating around COM is 2/5*m*r^2, or 2/5*I if I=m*r^2. An explanation to the previous question would be great: is it just convention when using the torque formula to just use MOI of the COM? Thanks! :)
     
  9. Dec 1, 2015 #8

    haruspex

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    You have to be consistent about the axis you use for MoI and the axis for torque.
    The 7/5MR2 is for the point of contact axis, so would be appropriate for a torque about that point. Which force has a moment about that point?
    The 2/5MR2 is for the CoM axis. Which force has a moment about that point?
     
  10. Dec 1, 2015 #9
    Haruspex: Wait....what has me confused is that the center of mass of the sphere is rolling along in a linear fashion, so the center of mass never moves...just rotates. And, in response to your question, the contact point on the axis has a force being applied to it...friction? And the center of mass has a force being applied...gravity?? Basically, just trying to make the most educated guess possible. Maybe I am overthinking it, it happens. Anyway, appreciate the response. Anything else would be great. I am still confused about this stuff, and whether I need to use the parallel axis theorem at all for this problem.
     
  11. Dec 1, 2015 #10

    haruspex

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    If the centre of mass moves (linearly), in what sense is it not moving?
    Those are true statements but not what I asked.
    Which force has a moment about the point of contact with the ground? Which force has a moment about the centre of mass?
    (I believe this is what your tutor was getting at.)
     
  12. Dec 1, 2015 #11
    Huh, I am really confused about your first question lol. In what sense is it not moving? I guess it is not moving up or down or in circles f I rotate my axis appropriately?
    And your second question...which force has a moment "about" the center of mass? I mean, the static force is pushing the bottom of the sphere in a tangential fashion. my best guess: The gravitational force has a "moment" about the center, because that is where the force is being applied? The static frictional force pushes the bottom, so it has its own separate "moment?" I am really confused, haruspex. Please elaborate more. Anything solid would be great :) Maybe I am misunderstanding the meaning of moment. Like I said, stabbing in the dark here. Haven't sliced through the curtain that is blocking the sun yet...haha so poetic.
     
  13. Dec 1, 2015 #12

    haruspex

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    It was you that wrote
    I'm asking you what you meant by that. It makes no sense to me.
    No, you seem to have a basic misunderstanding about moments.
    If a force ##\vec F## is applied at a point P which is at location ##\vec r## relative to an origin O then the moment of the force about O is given by the cross product ##\vec r \times \vec F##.
    There are various ways of expressing this relationship in scalars. One is: the magnitude of the moment about O of a force of magnitude F is F multiplied by the perpendicular distance from O to the line of action of F.
    See https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/ for a fuller discussion.
    In particular, note that a force has zero moment about any point that it passes through. (This is what your tutor meant about r being zero.)
    There are three forces acting on the sphere: gravity, normal force and friction.
    If you take moments about the centre of the sphere, the normal force and gravity both act through there so have zero moment. The only torque comes from friction: ##rF_{fric}=\tau_{CoM}##.
    If you take moments about the point of contact, the normal force and friction both act through there so have zero moment. The only torque comes from gravity: ##xF_{g}=\tau_{contact}##, where x is the horizontal distance from the point of contact to a vertical line through the centre of the sphere.
     
  14. Dec 1, 2015 #13
    Damnit, then I know what a moment is. I was referring to the moment of inertia, which from my understanding is not torque, but is a component of it by the torque equation torque= I*alpha. Anyway, I checked out your link! https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/
    Yes....I know that a force applied to a point on any object has no "moment" or "torque" about that particular point, if that is what you are talking about.
    What...if x is the distance from the point of contact to the center of the sphere, then the friction force would be the torque? Due to cross product? Sorry man, how does this relate to parallel axis theorem...?
    Also, see below.
    I need help determining the moment of inertia, which I believe is different from torque, for a rolling sphere. I understand that a force applied through the center of mass results in a zero moment/zero torque.
    Can you just tell me your opinion? I still feel the heuristic approach is getting me nowhere. So, 7/5mr^2 vs 2/5mr^2 for rolling sphere and why. Pretty frustrated right now, as I am more confused now than when I originally posted. Probably going to throw the towel temporarily in on this one.
     
  15. Dec 2, 2015 #14

    haruspex

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    Consider a rolling sphere and a point on the ground as a reference axis.
    There are two ways of finding the angular momentum of the sphere about the reference point:
    1. Using MoI about centre of sphere:
    This method considers the motion of the sphere as the sum of two motions, a linear motion of the mass centre and a rotation about that centre.
    The angular momentum of the rotation is 2/5mr2w; the angular momentum of the linear motion about a reference point on the ground is mvr = mr2w.
    Total = 7/5mr2w.
    2. Using parallel axis theorem:
    7/5mr2w.

    Likewise with torques and angular accelerations. You can use the MoI about the mass centre and the total torque about the mass centre (in this case, frictional force x radius), or you can use the MoI about point of contact with ground and the torque that gravity exerts about that point. These will produce the same answer, though you have to work through a few more steps to see it.
     
  16. Dec 2, 2015 #15
    If the origin for calculating torque at center of sphere,
    fR=I1ω
    If taken at point of contact,
    MgSinθR=I2ω

    f=2/7 MgSinθ
     
  17. Dec 2, 2015 #16
    Thanks everyone, I understand it much better now :)
     
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