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What will be the type of move and velocity of a sphere rolling without slipping on a flat surface?

  • In my mind friction force must be const. (there are no reasons for it to be variable). So epsilon is const.

  • so the linear acceleration of mass centre = epsilon*R

  • So linear acceleration of mass centre is const, and in the opposite direction of where the sphere is rolling.

  • Therefore the sphere will be rolling slower and slower and It'll stop at some time t.

I'm right? Could somebody explain me my mistakes?

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  • $\begingroup$ What is "epsilon"? $\endgroup$
    – NickD
    Commented Apr 24, 2017 at 21:47
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    $\begingroup$ I should have written "angular acceleration", sorry $\endgroup$ Commented Apr 24, 2017 at 21:50
  • $\begingroup$ Are you asking about the sphere moving up an incline or along a horizontal surface? If it is the latter then the frictional force is zero. $\endgroup$
    – Farcher
    Commented Apr 24, 2017 at 23:31
  • $\begingroup$ @Farcher, there is rolling friction associated with a "horizontal roll", but it is admittedly very low. $\endgroup$ Commented Apr 24, 2017 at 23:52
  • $\begingroup$ @DavidWhite I agree. I was try to find out whether to was a real life situation or a theoretical problem type situation. $\endgroup$
    – Farcher
    Commented Apr 25, 2017 at 5:06

2 Answers 2

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If the surface is horizontal, there is no rolling friction and there is no slipping between the sphere and the surface then there will be no frictional force acting and the centre of mass of the sphere will maintain a constant velocity and the angular velocity of the sphere will be constant.

The no slipping condition is $v_{\rm cm} = r \omega$ where $v_{\rm cm}$ is the translational speed of the centre of mass of the sphere, $r$ is the radius of the sphere and $\omega$ is the angular speed of the sphere.

There is no vertical movement because the net vertical force on the sphere is zero; the weight of the sphere is equal in magnitude and opposite in direction to the normal reaction of the surface on the sphere.

If a frictional force was present and it acted in the opposite direction to the motion of the sphere it would cause the centre of mass of the sphere to move slower in the horizontal direction but the torque about the centre of mass of the sphere produced by the frictional force would try and make the sphere rotate faster.
You cannot have the translational velocity decreasing whilst the rotational velocity was increasing and the no slipping condition maintained.
Hence there can be no frictional force in a direction which is opposite to the motion of the sphere.
A similar argument will show that there can be no frictional force in a direction which is in the same as the motion of the sphere.

Hence there is no frictional force acting on the sphere at it will maintain its constant translational and rotational motion forever.

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The sphere perform a constant speed movement. The friction no perform work because the sphere movement respect to the floor is upward but the sphere center of mass is moving perpendicular to the floor.

EDIT: Ok, look the picture. The black circumference is the sphere at some moment $t_0$ and the red one is at $t_1=t_0+\Delta t$.

The section marked in blue color is the part which have contact with the floor, if you follow these part, is moving upward for some later moment.

Imagine a xy plane (i forgot draw it) such that the positive y axis is directed to up. The friction force is acting over the blue section, so the force direction is to down $\vec{f}=f\,(-\hat{\mathbf{y}})$. The velocity of center of mass $\vec{v}_{cm}=v_{cm}\,\hat{\mathbf{x}}$ The work realized by a constant force is defined as follow, $$W=\vec{F} \cdot \vec{d}$$ Where $d$ is the displacement in the movement direction. This is the reason because the friction force doesn't make work, hence, the velocity the sphere is constant all over the flat surface.

Sphere

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  • $\begingroup$ I don't understand. " sphere movement respect to the floor is upward but the sphere center of mass is moving perpendicular to the floor." Shall you explain it to me, please? $\endgroup$ Commented Apr 25, 2017 at 20:01
  • $\begingroup$ ok, I 'll updete my answer. $\endgroup$ Commented Apr 26, 2017 at 0:12

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