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Rolling motion on inclined plane: centre of mass acceleration, friction, torque

  1. Aug 8, 2009 #1
    Hi,

    When a circular section object (cylinder, sphere of radius r) rolls down an inclined plane (at angle theta) it experiences both linear and angular acceleration due to gravity. I have read multiple times that:

    mg sin(theta) - friction = m a

    where friction = Torque/r

    From there on, everything is clear:

    moment of inertia * angular acceleration = friction * radius of ball = Torque

    a= angular acceleration * r

    and finally:

    a = (mg sin(theta)) / (m + moment of inertia * r^-2);

    friction = moment of inertia * a * r^-2

    But, going back to the first equation there is a question that has always profoundly bothered me:

    mg sin(theta) - friction = m a

    Why does friction appears here? Isn't it a force applied at distance r in the contact point with the surface and directed down, parallel to it?

    I understand that friction is needed to avoid slipping so it should exist. Moreover, that the weight must be reduced by some force to represent the final rotational energy. I mean, if we had just

    mg sin(theta) = m a

    and presume rotation then we would have the same final kinetic energy as a pure traslation with no friction (slipping) but with an impossible extra rotational energy.

    Summarizing, I recognize that:

    At centre of mass:
    mg sin(theta) - Unknown force = m a

    At contact point:
    Torque = moment of inertia * angular acceleration = friction * radius of ball

    But how do we get Unknown force = friction?

    Thanks in advance!
     
  2. jcsd
  3. Aug 8, 2009 #2

    Doc Al

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    Friction is directed up the incline, opposite to gravity. Yes, it is applied at a distance r from the center. So?

    I am unclear as to what's bothering you. Without friction, the object would slide without rolling. When friction acts it does two things: It exerts a torque and thus causes rolling; It exerts a force that reduces the translational acceleration.
     
  4. Aug 8, 2009 #3
    What bothers me is that I don't understand why this force, applied at a distance r from the center, reduces translational acceleration as if it were applied directly at the center of mass.
    (I feel I am missing something very fundamental : ) )

    Thanks for your prompt response!
     
  5. Aug 8, 2009 #4

    rcgldr

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    Note also that rolling involves static fricton. It's also possible for an object to be rotating and sliding at the same time, and with sufficient incline angle versus friction, the object continues to slide and never transitions into rolling.
     
  6. Aug 8, 2009 #5

    Doc Al

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    Yes, you are missing something fundamental. The translational acceleration of the center of mass of an object (given by Newton's 2nd law) depends only on the net force on the object. It does not depend on where the forces are applied on the object. (Of course rotational acceleration certainly depends on where the forces are applied.)
     
  7. Aug 8, 2009 #6
    Nice! That was what I was looking for... I think I will regard (rotational) dynamics from a completely different perspective now :shy:.

    Thank you very much for your help!
     
  8. Aug 8, 2009 #7
    Is this also a way to look at it:?

    Some of the pot. energy the object has lets say, at the top of the inclined plane, is turned into rotational kinetic energy because friction exists? So not all the energy goes into linear or translational kinetic energy. Therefore the acceleration at the center of mass cannot just be sin(theta)*g ???????

    sorry for butting in.
     
  9. Aug 8, 2009 #8

    Doc Al

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    Yes, perfectly correct.
     
  10. Aug 9, 2009 #9
    I posted this in another thread;

    The previous two posters are very correct. The principle outlined can be expressed mathamatically with out friction.

    In the rolling case for a hard solid ball of uniform density;

    m*g*h = rotational K.E + linear K.E

    mgh = (1/2)*I*w^2 + (1/2)*m*v^2

    I = moment of inertia
    w = angular velocity of ball
    v = linear speed
    r = radius of ball

    now a rolling ball of diameter r turning at frequency has a linear speed v of;
    v = 2*pi*r*f = w*r

    so w = v/r

    Therefore:
    mgh = (1/2)*I*w^2 + (1/2)*m*v^2

    becomes:+ m*g*cos(theta)*b/r)*d
    = (1/2)*((2/5)*m*r^2)*(v/r)^2 + (1/2)*m*v^2
    g*h = (1/2)*((2/5)*r^2)*(v/r)^2 + (1/2)*v^2
    = (7/10)*v^2

    v= sqrt((10/7)*g*h

    So if h = 1m, d = 1m
    v = 3.72 m/s
     
    Last edited: Aug 10, 2009
  11. Aug 9, 2009 #10

    rcgldr

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    Rolling resistance is related to the total force between surfaces, not just the normal force. If a car moving slowly enough to ignore aerodynamic drag is rolling down a hill at constant speed due to braking, or a high rolling resistance factor, then the total force at the tires is m g and not m g cos(θ).

    Only in the case of a frictionless inclined plane, will the total force exerted by the plane be equal to the normal force = m g cos(θ), as the the plane exerts zero force in the direction of the plane.
     
    Last edited: Aug 9, 2009
  12. Aug 9, 2009 #11
    I get what you are saying about the rolling resistance. What I don't get is the exact formulation of the total frictional force (not aerodynamic c drag) that the ball experiences. If I can do this for a ball it will work for a wheel or any other rolling object.

    Given I approached this problem from the principle of conservation of energy work must be done against friction. Given this is not a frictionless incline then I would need to include the frictional forces up the hill. Thus would the work done against friction up the hill be = 2*pi*r*f*N
    where N is the number of revolutions, r is the radius and f the static friction up the hill.
    Could this be right or am barking up the wrong tree again. How would rolling resistance then be added in? Could it be a direct addition of m*g*cos(theta)*b/r? Theta is included as it on a hill and the normal force to the plane will be reduced by cos(theta). I feel I am missing something.

    Thanks for pointing out my errors.
     
    Last edited: Aug 9, 2009
  13. Aug 9, 2009 #12

    Doc Al

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    If you ignore the complication of rolling friction and only worry about static friction, then there's no work done against friction and mechanical energy is conserved.
     
  14. Aug 10, 2009 #13

    rcgldr

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    No, because the rolling resistance is some function of the total force exerted by the plane onto the object, and this force is a function of the objects acceleration. If the object isn't accelerating, then the force from the plane equally opposes the force from gravity, so the force exerted by the plane onto the object = m g (in a vertical direction). In this case the force in the direction of the plane is m g sin(θ) uphill, and the force perpendicular to the plane is m g cos(θ). The other limit is if the plane is frictionless, in which case the force parallel to the plane is zero, while the force perpendicular to plane remains at m g cos(θ), so the total force is m g cos(θ).

    m g cos(θ) <= total_force <= m g

    In your simplified case, the force from rolling resistance = crr total_force, where crr is the coefficient of rolling resistance.
     
    Last edited: Aug 10, 2009
  15. Aug 10, 2009 #14
    So it variable. Thanks.
     
  16. Aug 10, 2009 #15

    rcgldr

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    It is variable, but you could just consider rolling resistance to be some constant force parallel to the inclined plane.

    linear force = gravitational component - friction force related to anguar accleration - rolling resistance

    define c = rolling resistant force / m
    define fa = surface (friction) force related to angular acceleration
    define fc = force related to rolling resistance = m c
    define fg = gravity force in direction of plane = m g sin(θ)

    m a = m g sin(θ) - m (Ia/Ih) a - m c
    a (1 + Ia/Ih) = g sin(θ) - c
    a = (Ih / (Ia + Ih)) (g sin(θ) - c)

    fa = m (Ia/Ih) a
    fa = m (Ia/Ih) (Ih / (Ia + Ih)) (g sin(θ) - c)
    fa = m (Ia / (Ia + Ih)) (g sin(θ) - c)
     
    Last edited: Aug 10, 2009
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