# Energy dissipation due to frictional force in rotational motion

Let me have a ramp with length l and height h. What I want is to roll that ball on the plane without slipping .

When I face this types of problem, I try to consider the rotational energy and kinetic energy in total. Therefore total energy $E= E_k + E_w(rotational)$ I see that a lot of problems have been solved by using the above formulae. But what about the dissipation of energy by frictional force? Don't we have to consider it?

Where you write 'roll that ball without slipping', that is a crucial point. If it wasn't for the friction force you've drawn in your diagram the ball would indeed be simply slipping (sliding) down the slope without any rotation going on.

It's that friction force that prevents slippage and that provides the torque around the centre of the ball: torque = friction force times radius ($r$) of the ball. This way the friction force provides the work that is converted to rotational energy.

What you indicated as $P_x$ provides the force for acceleration along the slope and the work it does is converted to translational energy.

Without slippage, the ball will rotate with frequency $f$ and is related to the translational speed $v$ via $v=2\pi rf$.

The total kinetic energy is $E_k=\frac{mv^2}{2}+\frac{I\omega^2}{2}$. In this simple case, $E_k=mgh$ with $h$ the height of the slope. All other things being equal a rolling ball will reach a lower final speed $v$ than one purely sliding (without any friction) because the same amount of energy ($mgh$) is now partitioned into rotational and translational energy.

That rotation has a reducing effect on the acceleration along the slope $a$ can also be shown as follows.

The equation of motion along the slope is given by:

$ma=mg\sin\alpha-F_f$.

The equation of motion for the rotation is:

$F_fr=I\dot{\omega}$.

With $v=2\pi fr=\omega r$, differentiate to $t$ and get $a=\dot{\omega}r$ or $\dot{\omega}=\frac{a}{r}$.

So that $F_fr^2=Ia$.

So we have two linear equations in $a$ and $F_f$ and we get:

$F_f=\frac{Ig\sin\alpha}{R^2+\frac{I}{m}}$.

But this is an ideal case: smooth ball on a smooth surface with just the right amount of friction to avoid slippage. In 'real world' cases there will be extra friction and this will do work that is irreversibly lost: for example a golf ball rolling down a grassy slope. Resistance by the grassy surface will cause some potential energy to be lost to friction work.

• Let's say $O$ is the instantaneous axis of rotation. Can we conserve the angular momentum of the rolling body about this point $O$ from a point $A$ on the slope to a point $B$? If yes, why? Isn't frictional force providing an external torque to the rolling body? May 6, 2021 at 15:28