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Question on Rotational Motion and Torque Theory

  1. Feb 28, 2016 #1
    IMG_1963.JPG IMG_1965.JPG IMG_1966.JPG IMG_1966.JPG IMG_1966.JPG 1. The problem statement, all variables and given/known data

    A cylinder of mass M and radius R rolls (without slipping) down an inclined plane whose incline angle with the horizontal is theta. Determine the acceleration of the cylinder's center of mass and the minimum coefficient of friction that will allow the cylinder to roll.

    2. Relevant equations

    Sigma Torque = moment of inertia * angular acceleration
    F = ma
    3. The attempt at a solution

    So I am trying to understand this problem and theory, and I am left wondering the following:

    1. Why is it that we can write F = ma equations using the same forces that are used in rotational motion? Mgsin(theta) - Ff = ma?
    2. Why do they say that the "static friction" supplies the torque needed to allow the cylinder roll smoothly?
    3. Why do they take the torque around the contact point to solve for acceleration using the moment of inertia (3*M*R^2)/2 instead of the regular MR^2/2?
    4. Why do they say that the gravitational force Mgsin(theta) is the only one causing the torque when they start doing the math, but say earlier that "static friction supplies the torque that allows the cylinder to roll smoothly?
    I have a physics test tomorrow and am frantically trying to understand these points. Any help is greatly appreciated.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Feb 28, 2016 #2
    i think you have a body which is rolling down an incline- in apure rolling there is no slipping means the two surfaces the body and the incline do not move relative to each other during motion so no kinetic friction will be operating- so any force equation should not involve a frictional term.
    so please correct your equation.
    you have the body as purely rollong means a line of contact is with the body and the body is turning about this line of contact ; what is the force that can rotate the body about this line of contact-
    (i) the available force is the weight of the body acting vertically downward
    (II) and the body is on the surface so a reaction of the inclined plane surface acting on the body- no third force is present - so why not resolve the force along the plane and perpendicular to the plane-as finally the motion is along the plane.
    then we come to situation that the body will have an acceleration along the plane- due to force Mg sin(theta)-
    now as it is rotating about the contact line -this rotation must have been produced by a force-the only force is the above one
    so take vector product of R and F ,they areat 90 degree so the torque = R.( Mgsin (theta)) = moment of Inertia about the axis of rotationX Angular acceleration; so use parallel axes theorem for I(moment of Inertia)
    so you get the acceleration.
    for getting the condition for the coefficient of friction such that slipping can start - one will have to consider the force of friction opposing the slipping motion .....
     
  4. Feb 28, 2016 #3
    Well... the force of friction is involved here. Here is a video of dan fullerton explaining this problem (but I still do not understand the points that I listed above) at time mark 17 minutes: . Notice he uses force of friction. Apparently, this is a standard problem in all colleges curriculums.

    So the friction is involved here. So we do have a force of friction and we have Mgsin(theta). So why are we using Mgsin(theta) in the torque and not force of friction?

    I understand how parallel axis theorem works and how they calculate the moment of inertia, but I think that the whole cylinder/disk/object is rotating around the center of mass when it goes down the incline, and not one of the ends.... well I do recall reading that instantaneously, the speed of the point on the object in contact with the plane is 0 m/s if the object rolls without slipping. Could that have something to do with that? And why do they say that the object rotates around a point on the circumference instead of around the center of mass????
     
  5. Feb 28, 2016 #4
    Fullerton starts the problem in the video at 13:49. My bad.
     
  6. Feb 28, 2016 #5
    If on a purely rolling case one applies the force of friction then the energy conservation can get affected as frictional forces can dissipate energy ...so that assumption can be tested experimentally
    2nd the body is turning about the line of contact and momentarily its velocity of the point of contact is zero- but rotation is taking place about the axis tangent to the surface so on what basis the moment of inertia will be calculated about the axis of the cylinder ,i fail to understand. as you are taking torque about that axis .

    if you are turning a body about an axis the point on the axis will have zero displacement .
    one is not saying that frictional forces are absent ; however the frictional force F(friction) = Coefficient X Normal reaction must be present but its just preventing the body not to slip and its torque about axis of rotation should be zero.
    consult say Resnick and Halliday's Principles of Physics.
     
  7. Feb 28, 2016 #6
    So why then is the expression for torque written with Mgsin(theta) instead of friction force?
     
  8. Feb 28, 2016 #7
    If you are taking the torque by the component of weight mgsin (theta) -turning the cylinder about the line of contact then the Moment of Inetia comes to be different and if you take torque by friction force then its about axis of the cylinder - the idea behind taking the mgsin(theta) term is perhaps the external force rotating the body as frictional forces generate after a tendency of motion is there- i think both ways should lead to same physical results -one can check.
    why i am talking of tendency of motion of bodies- is common experience that unless we start application of force the frictional forces can not be defined a priory-in both translational as well as rotational motion.
     
  9. Feb 28, 2016 #8
    Let me try to restate what you said and see if I understand what you are saying correctly:

    The frictional force results only AFTER the object starts turning. In this instance, the object starts to turn around the point of contact thanks to the mgsin(theta) force and only after the object starts turning do we have the friction force.... right?
     
  10. Feb 28, 2016 #9
    So thus, the friction force is not included in the torque as it is a result of the torque caused by mgsin(theta). Yes?
     
  11. Feb 28, 2016 #10
    You can indeed use torque about the C.M. but then you have to eliminate the force of friction since
    the torque = f R.
    Also, M a = M g sin theta - f for the acceleration of the C.M.
    Using torque about the point of contact s just a little more concise,
     
  12. Feb 28, 2016 #11
    I understand how the f=ma equation would work with mgsin(theta) - friction force.

    But.... if I am using torque around the contact point, it is because the contact point's velocity is 0 m/s... yes/no?

    And the friction force is a result of the object's movement around the contact point..... so the friction force does not cause torque, but the mgsin(theta).... so then the gravity IS included in the torque expression, but the friction force is not. Am I right in this interpretation?
     
  13. Feb 28, 2016 #12
    Obviously, the velocity of the contact point is not zero.
    Also, the force of friction exerted at the contact point cannot exert any torque about the contact point.
    You seem to be having trouble with the fact the angular velocity about the contact point is
    the same as the angular velocity about the C.M.
    Try calculating the speed of a point on the top of the cylinder (perpendicular to the plane);
    maybe that will clear up that point.
     
  14. Feb 28, 2016 #13
    I do not think conseptually one can put frictional force to be zero- as you start turning the frictional force starts to build up from zero and you mount your force higher it grows to a limit of static friction -it does not need motion-only a tendency to move-for example you keep a cylinder on a plane horizontal plank and slowly raise it to an incination say small angle theta(0) the cylinder may not roll- the torque by force along the plane may not be able to rotate the cylinder about the line of contact/nor even slide it -what is preventing the motion- the frictional force acting opposite to the tendency to move. that depends on the nature of surfaces in contact- when we take the torque by force along the plane- about line of contact the torque of frictional force about that axis is zero.(well i think that is the picture i draw up).
     
  15. Feb 28, 2016 #14
    Here is what I mean by the velocity equalling zero. Look at the third diagram. This comes out of the Jewett Textbook.

    So if we have a combination of rotational and translational motion... then the velocity at the contact point is zero. So this is where I think I have an instantaneously fixed point and the object is rotating around the contact point P.

    Thus, there is no torque by friction , but there is a torque by mgsin(theta). Right?
     

    Attached Files:

  16. Feb 28, 2016 #15

    haruspex

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    Moments, torques and angular momentum are (generally) only defined in respect of a given axis.
    You do have to be a little careful about choice of axis for some purposes, but it is always safe to take:
    • An axis through the mass centre of the rigid body
    • An axis through a fixed point in space
    In the present problem, you can take an axis through the mass centre of the cylinder, or one through the instantaneous point of contact with the plane. Both methods will yield the correct answer.

    The advantage of using the point of contact is that it eliminates torque due to friction. The moment of inertia about that point is, by the parallel axis theorem, 3mr2/2. If you don't need to find the friction, this is less algebra.

    If you take moments about the mass centre, the equation will involve the frictional force, so you have to combine it with the linear acceleration equation and the rolling condition equation (v=rω) to get the solution.

    As to which force causes the angular acceleration, again, it depends on the axis. If a block is sliding and accelerating down a ramp, it has angular acceleration about an axis on the surface of the ramp, though you would not normally think of it as rotating. If we interpret the question in the natural way, i.e. in terms of angular acceleration about the mass centre, then it is clear that the friction cause the angular acceleration, not the mg or the normal force.
     
  17. Feb 28, 2016 #16
    So if I have defined my axis to be at the center of mass, then my torque expression will include friction AND force of gravity (mgsin(theta))?

    Thanks for the help
     
  18. Feb 28, 2016 #17
    So assuming that my axis as at the center of the mass, I would set up my expression to be the following (using the friction force and the normal forces)

    R ( Mgsin(theta) + mu * Mgcos(theta) ) = [ (MR^2)/2 ] * angular acceleration.

    My two forces they are added in the left hand parenthesies, yes?
     
  19. Feb 28, 2016 #18
    All right. Here's what I just did

    R (Mgsin(theta) + mu * mgcos(theta ) ) = I *angular acceleration
    R(Mgsin(theta) + mu* mgcos(theta) ) = [MR^2/2] * [ a /R ] (my a is linear acceleration)
    Mgsin(theta) + muMgcos(theta) = Ma/2
    gsin(theta) + mugcos (theta) = a/2 [will return to later]


    OK I got the expression above. Now, let's use F = ma.... to get the following
    Mgsin(theta) - muMgcos(theta) = Ma
    gsin(theta) - mu
    gcos(theta) = a. [will return to it in the next line]


    OK. I have two equations that are marked with square brackets. Let's add them!

    gsin(theta) + mugcos (theta) = a/2
    + gsin(theta) - muMgcos(theta) = a
    --------------------------------------------------
    2gsin(theta) = 3a/2

    a = 4gsin(theta) / 3.

    Plug in the a into the equation: gsin(theta) - mugcos(theta) = a = 4gsin(theta) / 3 [second equation]

    I solved the expression for my mu and I got the result to be -tan(theta)/3 = mu.

    My book's solution has the correct answer to be (positive):
    tan(theta)/3 = mu

    Where did I mess up? I cam in really close
     
  20. Feb 28, 2016 #19

    haruspex

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    No. The torque about a given axis due to a given force is the vector product of the force and the displacement from the axis to the point of application of the force. In scalar terms, it's the magnitude of the force multiplied by the perpendicular distance from the axis to the line of action of the force. See https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/.
    The normal force and the gravitational force both act through the mass centre, so or them the perpendicular distance is zero, and they have no moment about the mass centre.
    Likewise, the frictional force and the normal force have no moment about the contact point.
     
  21. Feb 28, 2016 #20
    Ahh.... I see. So then my mgsin(theta) would not be included but my friction force would be included in the torque expression assuming my axis of rotation is at center of mass
     
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