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When we push a sphere on a rough horizontal surface, it slows down because there is a net torque on it (by normal force and friction acting in opposite directions) and this causes its angular speed to decrease. But work done by all the forces on it is zero too, so why does it's kinetic energy decrease?

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    $\begingroup$ Related: physics.stackexchange.com/q/149409/305718 $\endgroup$
    – ACB
    Aug 25, 2021 at 6:32
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    $\begingroup$ The work done isn't zero. Friction is applying a force over some distance. Just that the direction of movement isn't the same direction as the force as you are typically used to. $\endgroup$
    – DKNguyen
    Aug 25, 2021 at 16:03
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    $\begingroup$ I am unsure about the scenario you are considering: 1) The normal force and friction do not act in opposite but orthogonal directions. 2) Is the sphere rolling at the beginning or not? If no, its angular speed increases, while it slows down (this is a typical rolling-with-slipping problem). If yes, it would not slow down due to kinetic friction. Please edit your question to clarify. $\endgroup$
    – Wrzlprmft
    Aug 26, 2021 at 12:02
  • $\begingroup$ The torque of friction and normal force act in opposite direction and net torque of these two act in opposite to the direction of angular velocity which causes angular velocity to decrease. Hope you understood it now. $\endgroup$
    – Mr. Wayne
    Aug 26, 2021 at 12:11
  • $\begingroup$ @Mr.Wayne: I am not asking you to explain the physics of the scenario; I am asking you to clearly describe your scenario, in particular whether your sphere is initially rolling or not. — The torque of friction and normal force act in opposite direction – Comparing the direction of a force and torque (its axis) makes little sense, but even then this statement is not correct: The normal force acts in vertical direction; the frictional force acts in horizontal direction; the axis of the torque caused by the latter is also horizontal. All three are orthogonal, not opposite to each other. $\endgroup$
    – Wrzlprmft
    Aug 26, 2021 at 12:29

6 Answers 6

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Even if you assume a perfectly frictionless surface, the ball would still slow down because of inelastic deformation.

Whenever a force is applied to an object it deforms, and when the force is removed, the object returns to its shape. In the case of rolling ball, both the point of the ball and the ground keep deforming and returning back.

This leads to continuous loss of energy. This is also why a hard ball will roll for longer than a soft ball which deforms much more.

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    $\begingroup$ A very visual example of this would be trying to roll a water balloon. You'll be able to see the point at which the deformation of the balloon causes the 'ball' to stop rolling. Plus there's a chance it might just pop, which would be a fun lesson on checking your modelling assumptions! $\endgroup$
    – Joe Bloggs
    Aug 25, 2021 at 15:17
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    $\begingroup$ Wait, you are saying that friction is, in fact, not zero ;-) ? $\endgroup$ Aug 25, 2021 at 15:51
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    $\begingroup$ On a perfectly frictionless surface, it wouldn’t roll at all (unless given an initial spin when it started, but that would still be independent of its translation). For a rolling object, the ideal case isn’t frictionless, but rather “perfect friction,” no slipping. $\endgroup$
    – KRyan
    Aug 26, 2021 at 15:45
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    $\begingroup$ @KRyan You can see an approximation of that situation with bowling balls. $\endgroup$
    – Herohtar
    Aug 26, 2021 at 20:09
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You are confusing the ideal case with the non-ideal case.

  • Ideal scenario: No elastic deformation (so no internal work done), no air resistance, no uneven normal forces due to a flexible or soft surface that can cause counteracting torques, no kinetic friction forces. In this scenario, nothing does work and no energy is lost. The kinetic energy will remain and the rolling speed the wheel has, both linear speed of centre-of-mass as well as angular velocity, will stay constant and unchanged.

  • Non-ideal scenario: When any of the above mentioned effect do take place, then they can cause heat loss (energy lost to e.g. the air or to frictions) as well as negative work done against the motion.

In other words, if you do have counteracting torques and other non-ideal then there is being done negative work. This is the reason for the wheel slowing down. In the ideal scenario, the wheel would never stop (on horizontal ground).

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A rolling sphere slows down only when you stop applying the torque (basically when you instantaneously push a sphere forward rather than constantly applying the torque).

In the former case, once you have applied the instantaneous torque on a sphere, it will experience somewhat of a counter torque as its rolling due to the irregular surface of the sphere. This is due to the fact that in a deformed rolling sphere, the Normal forces from ground donot exactly pass through the centre of mass.

Counter Torque on a rolling sphere

Due to this counter torque, the sphere slows down. This also means that if you have a hypothetical sphere which has no deformation or irregularities, then it would not experience any counter torque (since all normal forces would pass exactly through centre of mass) and would roll forever. Also note that static and kinetic friction has no role in slowing a perfectly rolling sphere.

In the latter case, since you are constantly applying a torque, there the net torque remains constant.

$$\tau_{net} = \tau_{applied} - \tau_{counter} = constant$$

Since the net torque is constant and positive, so ball rolls forever.

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Some review just to be sure I hit whatever the gap is. Bear with this:

You have three friction coefficients: static, kinetic, and rolling

Kinetic

A sphere can slide if the force required to angularly accelerate it enough to make it roll exceeds $N$ times the appropriate coefficient (kinetic or static to start depending on the situation, and then kinetic).

Static

If the friction is enough to make it roll then that is the exact tangential force provided. And it does affect translation (ie by keeping it rolling). But that tangent aspect does no work as it acts on a still point ($F \cdot x=0F$), and it does not lower energy as you mentioned.

Rolling

If rolling, you also have rolling friction. It is providing a retarding horizontal force, at the center of the wheel. Just apply the force there. It does that in the real world by the normal force not being exactly vertical, but still pointing radially, at the axis, because the surface that it’s riding on will indent a little, so it’s like riding over a super low speed-bump. (The wheel deforms also but this has less effect; that’s why I try to emphasize focusing on the surface; these are elastic deformations.)

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it slows down because there is a net torque on it (by normal force and friction acting in opposite directions)

The normal force doesn't supply a torque. It's normal. And if the ball is rolling on a flat surface, in the ideal case friction doesn't change the angular momentum, since it is applied on at the point of contact, and the point of contact is not moving, so the torque is not being applied over any distance.

In practice, there are various energy losses. The very fact that you can hear a ball rolling means that energy is being dissipated in vibrations. The ball will be deformed, the surface isn't perfectly flat, etc.

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If the ball is subject only to external forces coming from the floor (plus obviously its weight), it may slow down resulting in a decrease in its kinetic energy only by work done by friction forces and friction torque. Let's now for simplicity consider an ideal case with no rolling resistance, then there can be work done by tangential friction only if the ball slippers or slides, whenever the ball moves in "pure" rolling (i.e., contact point has instantly zero speed) the tangential friction can no more be dynamic and can only be static, with null work; anyway, be careful that a not null static friction generates a not null torque consuming the kinetic energy of the ball. The equilibrium for this exemplum is reached when the tangential forces are off and the ball translates and rolls with an angular velocity equal to the ratio of the speed of its center to the length of its radius: $$\vec{\omega}=\vec{v_0}\times\vec{r}\quad.$$ If $\omega=0$ then the friction torque will have consumed all the kinetic ball energy and the ball will stop.

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