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Friction on spinning object

  1. Oct 2, 2009 #1

    dws

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    I've been wondering for some time and it seems I still have doubt on how friction works on spinning object.

    Assume a situation like this (top down view):
    http://img33.imageshack.us/img33/5889/physics.jpg [Broken]

    Assume dimension of the box is w and h, ignore the thickness of the box for time being. Uniform density, so center of mass is in the center of the box. And two forces F are applied just like in the image.
    [tex]\sum F = 0[/tex], so no translational movement occurs, it will be pure spinning motion.

    [tex]\sum \tau = F . w /2 + F . w /2 = F.w [/tex], but that is if we assume no friction.

    My question is how to compute the friction in this case?
    My guess is [tex] F_{f} = \mu mg [/tex] like usual with the direction the opposite of F.
    So new torque equation will be:
    [tex]\sum \tau = F . w /2 + F . w /2 - F_{f} . w/2 - F_{f} . w/2= (F - F_{f}).w [/tex]

    But is it really correct? I have a doubt in which direction the friction force is applied, because it can make a different resulting torque due to different perpendicular distance to the center of mass.

    Anyone can enlighten me? Thanks in advance
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 3, 2009 #2

    Cleonis

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    You're not giving enough information. You are making me guess what it is that worries you.

    If the spinning object is symmetric with respect to its center of mass, then air friction will be a torque around the same axis as the spinning.

    If the torque that spins up the object remains constant, then as the spinning rate increases the air friction increases accordingly. Eventually a point will be reached where the torque from air friction opposes the accelerating torque exactly, at which point acceleration ceases.

    That tells you the torque from the air friction at that particular spinning rate, and no more. Of course the air friction force is in fact distributed over the entire surface of the object. You have no details of that distribution, all you have is the resultant torque.

    Cleonis
     
  4. Oct 3, 2009 #3

    dws

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    Hi sorry for not being clear. I'm now concerned about the kinetic friction between object and the ground not the drag friction caused by the air. Because when the object spins, it 'slides' on the ground, therefore it should have kinetic friction as well. But I'm not sure how can I draw this kinetic friction on the image I put. My guess it should be similar to your explanation about the air drag that it should be distributed to the entire surface touching the ground. But then how can I calculate the torque caused by the kinetic friction?

    Thanks
     
  5. Oct 3, 2009 #4
    Assuming the dimension h to be small(i.e assuming it to be a long bar of negligible width) , you could divide up the whole block into infinitesimal elements of dimensions dx and h , each at a distance x from the axis of rotation, then a frictional force dF would act on each element. You could compute the mass of each element by assuming M as the mass per unit length( not area as i have neglected the dimension h).Then the mass of each element would be Mdx , and the frictional force dF on each element would be [tex]\mu[/tex]Mdxg. The torque due to friction of each element would then be [tex]\mu[/tex]Mdxg*x =[tex]\mu[/tex]Mgxdx, Thus we have ,
    dT=[tex]\mu[/tex]Mgxdx. Therefore T=[tex]\int[/tex][tex]\mu[/tex]Mgxdx=Mgw[tex]^{2}[/tex]
    Integrating within the limits of -w and w.Now as the entire block could be tought of as composed of say n such bars , and a torque of Mgw[tex]^{2}[/tex] acts on each bar.Thus a torque of nMgw[tex]^{2}[/tex] acts on the entire block = M'gw[tex]^{2}[/tex](where M' = mass per unit length of the block)This torque would have to be subtracted from the torque due to the two F's in your figure .I hope this clears your doubt as to which direction the force is applied. Note that the frictional force does not act at a particular point from the axis of rotation , nor would it be feasible to assume that it acts at a particular point. The force is uniformly distributed about the body , and this force distribution is determined from the mass distribution.Since the mass distribution is uniform , the methods of integral calculus greatly simplify the problem.
     
    Last edited: Oct 3, 2009
  6. Oct 3, 2009 #5

    arildno

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    This is how you should proceed:

    1. Let the sides be 2b and 2a, the height of the box h, M its mass.

    2. We will assume a velocity-independent frictional force that at each point is proportional to the normal force required for the column above to stay vertically at rest; any stresses in the vertical direction are ignored so that this is simply the weight of the column.

    3. Since the object is merely spinning around the C.M, it follows that each tiny area element traverses a circle.
    The frictional force at such a point is therefore along the tangent of the circular path traversed.

    4. It therefore makes sense to introduce polar coordinates, giving:
    Differential normal force:
    [tex]dn=gdm=\rho{gh}rdrd\theta,\rho=\frac{M}{4hab}[/tex]

    Differential frictional force df:
    [tex]df=\mu{dn}[/tex]

    Differential torque with respect to central axis:
    [tex]d\tau=rdf=\mu\rho{gh}r^{2}drd\theta[/tex]

    Thus, the total torque to the force of friction is given by:
    [tex]\tau=\mu\rho{gh}\int_{A}r^{2}drd\theta[/itex]

    "A" being the rectangular bottom.

    That integral is somewhat tricky to solve, maybe I'll post the result later.
     
  7. Oct 3, 2009 #6

    arildno

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    In order to solve the integral, we note that the four sub-rectangles meeting at the C.M, having sides "a" and "b" will contribute equally. Thus, we limit ourselvs to one such sub-rectangle, and multiply our final result with..4

    Furthermore, we divide that sub-rectangle into the two congruent triangles, rotated about the diagonal of the sub-rectangle.

    We only calculate the result for one of those triangles; the contribution from the other will be gained by simply switching the roles "a" and "b" play.

    We let "a" be the length in the x-direction, "b" in the "y"-direction.

    Since we have:
    [tex]0\leq{x}\leq{a}[/tex]
    we get:
    [tex]0\leq{r}\leq\frac{a}{\cos\theta}[/tex]
    and:
    [tex]0\leq\theta\leq\cos^{-1}(\frac{a}{\sqrt{a^{2}+b^{2}}})[/tex]

    Thus, we get, from the one triangle:
    [tex]\tau_{triangle}=\frac{a^{3}\mu{gh}}{3}\int_{0}^{\cos^{-1}(\frac{a}{\sqrt{a^{2}+b^{2}}})}\frac{d\theta}{\cos^{3}\theta}[/tex]

    Now, we have the anti-derivative:
    [tex]\int\frac{d\theta}{\cos^{3}\theta}=\frac{1}{2}(\frac{\tan\theta}{\cos\theta}+\ln(\frac{1+\sin\theta}{\cos\theta})}[/tex]

    Now, inserting the limits, simplify a bit, and adding together the torques of the 2 triangles, (remembering to switch "a"'s and "b"'s" for the other one), and multiplying with 4, we get the total torque:
    [tex]\tau=\frac{\mu{gM}}{3}(R+\frac{a^{2}}{2b}\ln(\frac{R+b}{a})+\frac{b^{2}}{2a}\ln(\frac{R+a}{b})), R=\sqrt{a^{2}+b^{2}}[/tex]
     
    Last edited: Oct 3, 2009
  8. Oct 4, 2009 #7

    dws

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    wow, thanks a lot for the answers.

    I somehow can understand the concept behind it, I will ponder and 'munch' on the actual equation at my own pace. The concept itself is much more important. I have been consulting with my friends in this field as well. It seems for many years I have slightly wrong concept of friction.

    once again, thanks :)
     
  9. Oct 4, 2009 #8

    arildno

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    To help you munch more efficiently:

    Kinetic friction at some point is ALWAYS in the opposite direction of that point's velocity.

    If an object, therefore, spins, whether in addition with linear motion or alone, then we cannot assign ONE direction along which the frictional force can be said to act.

    This is because different contact points on the object will have different velocities, and therefore, we can only speak of the direction of the frictional force as a local property, rather than a global property.
     
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