I'm a little confused.

Translational energy and rotational energy add separately, according to my textbook, to give the total kinetic energy of a moving object.

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That means that for a disk rolling without slipping at a certain velocity, the total kinetic energy would be:

enter image description here

Now, if its a hollow disk, the moment of inertia is:

$I = MR^2$

Giving us a total Kinetic Energy of:

$K = MV^2$

That is twice the kinetic energy the disk would have if it was slipping on a frictionless icy surface.

But this leads to some weird results that I can't make sense of.

For example, say both disks, the one slipping, and the one rolling without slipping, are moving at the same velocity side by side, and a strong wind starts blowing, pushing them with a force $F$ opposite to their direction of motion.

Using energy conservation, the rolling disk would go twice as far as the sliding disk before it comes to a stop, since it had double the initial Kinetic energy.

Since the disks had the same initial velocity, that means the rolling disk must've had half the deceleration as the slipping disk, even though they were moving against the same force and had the same mass!


Thank you!

  • $\begingroup$ I made amistake, I was thinking of the case when the disc that is forced to rotate (like the wheel of a car) in the direction of rolling, in which case the friction is in the forward direction, because at the point of contact the wheel is moving backwards. I erased my answer $\endgroup$
    – user65081
    Commented Nov 20, 2018 at 22:39

2 Answers 2


To make the maths simpler to understand, replace the wind by an applied force $F$ at the center of the disk, and start with the disk at rest.

In the no-friction case, the acceleration is simply $a = F/M$ where $M$ is the mass of the disk.

If the disk is rolling without slipping and has a linear acceleration $a$, it must also have an angular acceleration $\dot\omega = a/r$.

To produce the angular acceleration, there must be a friction force $F'$ at the ground producing a torque of $I\dot\omega$, so $F'r = Mr^2(a/r)$ or $F' = Ma$.

The resultant horizontal force on the disk is $F - F'$, so Newton's second law gives $F-F' = Ma$.

Substituting for $F'$, we get $a = F/(2M)$ which is the same result that you got using energy.

  • $\begingroup$ Just writing this for myself because I'm looking over old posts: in short, the ground exerts a force opposite to the applied force in order for the object to keep on rolling without slipping. If the object is a hollow cylinder, that force would be equal to the applied force, and thus both an acceleration or a deceleration gets halved for any applied force. $\endgroup$ Commented Dec 28, 2018 at 19:36
  • $\begingroup$ I can't be bothered to check this answer over but I trust that it was an excellent answer. I was going to write an answer explaining how the math shows that the total kinetic energy of a rolling sphere is the sum of the kinetic energy it would have if it had only the translational part and the kinetic energy it would have if it had only the rotational part. Now that I'm old enough, I don't have strong desires for other things like reputation points and its kind of nicer naturally not having really strong desires for them. Now my real natural desire is smartness and making sense of things. It $\endgroup$
    – Timothy
    Commented Oct 24, 2019 at 16:25
  • $\begingroup$ really is more preferable to myself to acknowledge the usefulness of this answer than to write an answer of my own. It's my passion to not accept as proof what the laws really predict that a sphere will do if the proof uses energy and ask myself how it can be proven in a more straight forward way without using energy such as using angular momentum. My answer would have proven a property of energy because I thought that was what the author was asking but it wouldn't have shown why a sphere is predicted to have the behaviour it has when a force is exerted on it. I really do like this answer $\endgroup$
    – Timothy
    Commented Oct 24, 2019 at 16:32
  • $\begingroup$ better than the one I was going to write myself. $\endgroup$
    – Timothy
    Commented Oct 24, 2019 at 16:32

For those who do not feel comfortable with math formulas:

Imagine spinning a wheel in the air, then drop it to the ground. It will start moving forward. The angular momentum, the rotational inertia will be the driving force pushing the wheel forward.

The same "driving force" exists in your spinning wheel that helps it against the wind.


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