# Rolling Without Slipping and Rotational Energy

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.

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

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!

Why?

Thank you!

• 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 – Wolphram jonny Nov 20 '18 at 22:39

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.

• 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. – Joshua Ronis Dec 28 '18 at 19:36

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.