1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rolling without slipping.

  1. Aug 3, 2013 #1
    The below questions are NOT HW questions, but I have a big exam coming and I would wholeheartedly appreciate some/any assistance with the following two issues, namely:

    (1) Direction of static friction in case of rolloing without slipping
    (2) Direction of α and VCM in case of rolling without slipping

    (1) Let us please examine the three diagrams attached:
    Case 1, showing a sloped plane, the cylinder is rolling down the plane, clockwise, and the static friction is causing a moment (with respect to the center), equal to fR (-z) causing it to roll. Here, the static friction is aiding the rolling.
    Case 2, showing a cylinder let's say, rolling without slipping, clockwise, whilst the (static?) friction, in the direction as delineated in the diagarm (I didn't draw it!), is creating a moment opposing that of the tension, causing the cylinder to slow down. Here, the friction is resisting the rolling.
    Case 3, supposing the cylinder is rolling at a given moment to the left (i.e. counterclockwise), which was defined to be the positive X direction, the static friction acting on the cylinder at the point of contact with the bar, would be to the right. Thus, the static friction is, again, aiding the rolling.
    Now comes my question - based on these three examples, I see that at times the static friction aids the rolling, whereas it could, on other occasions, resist it (case 2). How may I then determine its direction? How may I know whether it is to assist the rolling or act against it??
    (2) I have seen that, at times, VCM = -ωR (instead of ωR), in rolling without slipping. Why is that and, most importantly, how may I determine the correct sign for my equations??

    Attached Files:

  2. jcsd
  3. Aug 3, 2013 #2


    User Avatar
    Homework Helper

    The static friction opposes the relative motion of the surfaces in contact.
    When the cylinder rolls down the slope, it would slip downward without (enough) static friction. So the force of friction has to point upward along the slope to prevent it. With respect to the centre, it produces torque that rotates the cylinder clockwise. Clockwise rolling moves the CM down the slope, and the distance covered by the CM is equal to the length of trace of the cylinder on the slope, which is R*angle of turn. In unit time v=Rω.

    (The torque is vector product of the vector r (pointing from the point of reference to the point of application of the force) with the force. τ=rxfs.
    If the x axis points downward along the slope, the y axis is perpendicularly upward and the z axis is out of the plane, in the direction, from where you see the figure. The torque of the static friction is negative, so is the angular acceleration, and the resulting angular velocity. At the same time, the linear velocity v is the cross product [STRIKE]rxω=[/STRIKE] v=ωxr. The point of contact moves in negative direction with respect to the CM, but the CM moves in the positive x direction with respect to the point of contact: from here, VCM=-Rω. )

    The other answers come in subsequent posts.

    Last edited: Aug 3, 2013
  4. Aug 3, 2013 #3
    I thought v=ωXr. Moreover, based on your explanation, won't the point of contact (with a plane below the object) always move in the opposite direction to the direction of CM? Apparently, I must be getting it wrong still :-(.
  5. Aug 3, 2013 #4


    User Avatar
    Homework Helper

    In case 2, it is difficult to say the direction of the static friction. The tension applied at the top of the cylinder rotates it, and at the same time, accelerates the CM. If the cylinder slips, you can not say in what direction the surface of the cylinder moves with respect to the ground. Here it is better to write up the torque equation with respect to the instantaneous axis: τ=2RT, and I=1/2 MR2+MR2=3/2 MR2 now. 2RT=(3/2 MR2)α, and Rα=aCM, so aCM=4/3 T/M.

    As the acceleration of the CM is determined by the sum of forces, T+fs =M (4/3)T/M=4/3 T. That is, fs=(1/3) T, and points in the same direction as T.

  6. Aug 3, 2013 #5


    User Avatar
    Homework Helper

    Sorry, I mistyped it. v=ωXr. Yes, the point of contact moves with respect to the CM in opposite direction as the CM moves with respect to the point of contact.
    You do not need to use the vector notation. The direction of motion of the CM and the direction of rotation are quite straightforward. Use the scalar equation Vcm=Rω, with R the radius.

    Using the instantaneous axis as point of reference, you do not need to decide anything about the direction of friction.

  7. Aug 3, 2013 #6
    Okay, but supposing I did want to determine the relation between the sign of the velocity of center of mass and the angular velocity under rolling without slipping, and also the sign of the angular acceleration with respect to the ratio of the center of mass acceleration and and the radius, how do I go about it? I still haven't quite understood that one. Would you care to try and explain once more this delicate point, making particular reference to the third diagram I attached. It would also prove beneficial should you please comment on the direction of friction there as well. Thanks in advance!
  8. Aug 3, 2013 #7
    And I shall still have to determine the direction of friction for the equation of the forces.
  9. Aug 3, 2013 #8


    User Avatar
    Science Advisor
    Homework Helper

    hi peripatein! :smile:

    depends whether it's a driving wheel or a non-driving wheel

    a driving wheel is one with an applied torque (ignoring the friction) about the centre of rotation …

    direction of friction is the same as linear acceleration for a rolling driving wheel, but opposite for a rolling non-driving wheel :wink:

    (eg the friction from the road on the driving or braking wheels of a car is in the same direction as the acceleration or braking, but the friction on the non-driving or non-braking wheels of a car is in the opposite direction)
  10. Aug 3, 2013 #9


    User Avatar
    Homework Helper

    You can determine the angular acceleration from the torque with respect to the point of contact. In that case, you do not need to know the force of friction. After determining the acceleration, you get the force of friction from the equation maCM=ƩF.

    In case 3), if the cylinder rolls to the left somehow, and there is no other force but friction and the force of the spring, and the bar does not slip on the cylinder, the spring must be compressed, exerting a leftward force on the bar. The cylinder is accelerated by the force the bar exerts on it, and that is the static friction. The static friction exerted on the cylinder points to the left, accelerating rolling. The static friction exerted on the rod points to the right.
    You can write two equations, one for the acceleration of the rod, the other one for the angular acceleration of the cylinder. Calculate torque and moment of inertia with respect to the instantaneous axis.

  11. Aug 3, 2013 #10
    Newton's Second Law is written vectorially. You still have to determine the direction of the friction force.
  12. Aug 3, 2013 #11
    hi peripatein :smile:

    A simple way to find the direction of friction .

    Suppose a body(cylinder,sphere,ring etc) of mass M and radius R is resting on a horizontal plane with sufficient friction to prevent slipping.Now a force of constant magnitude F is applied at the top of the body rightwards.

    The key thing is to find how a and αR are related.

    First assume there is no friction present and find the relation between a and αR

    For translatory motion ,

    F=Ma (1)

    For rotational motion

    FR=Iα (2)

    Equating the two ,


    Case I:For solid cylinder I=MR2/2

    a=αR/2 i.e a<αR

    Now for the body to roll without slipping a = αR .Hence friction acts in rightwards direction to accelerate the body and at the same time providing anticlockwise torque to reduce angular acceleration such that a = αR is achieved.

    Case I:For hollow cylinder I=MR2


    In this case the desired condition a=αR is achieved without friction.Hence zero friction in this case.i.e the hollow cylinder doesnt require friction to start rolling without slipping.

    Similarly you can check other cases of solid and hollow spheres ,ring etc.The same approach works with rolling bodies on inclined planes.

    Hope that helps .
  13. Aug 3, 2013 #12


    User Avatar
    Homework Helper

    You may determine the direction of the friction force before determining the acceleration, or after it.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted