Energy Conservation in Rolling without Slipping Scenario 
A solid ball with mass $M$ and radius $R$ is placed on a table and given a sharp impulse so that its center of mass initially moves with velocity $v_o$, with no rolling. The ball has a friction coefficient (both kinetic and static) $μ$ with the table. How far does the ball travel before it starts rolling without slipping?

The solution I found starts by setting up a conservation of energy and setting $v = rw$:
$$ \ \frac{1}{2}m v_o^2 = \frac{1}{2}m v^2 + \frac{1}{2}Iw^2 \to v_o^2 = \frac{7}{5}v^2 \quad{(1)}$$
It then goes on to say $W = \Delta K_{rotation}$ and solves for $D$ :
$$  \ \int_{0}^{D} F_{f} dx = μmgD=  \frac{1}{2}Iw^2\quad{(2)}$$
There are a couple of things I do not understand about this approach. How does $(1)$ account for the loss of energy due to the friction force which is causing the rotation and the slipping that occurs before it starts rolling purely? Second, how does $(2)$ account for the change in center of mass velocity? Wouldn't $W = \Delta K_{rotation} + \Delta K_{transitional}$ ? I am most likely misunderstanding something and help is greatly appreciated.
 A: I am not sure about this solution. I would set up the equations of motion as follows.
Translational motion:
there is only one force acting on the system, which is dynamic friction of modulus $F_d=\mu mg$. The motion is uniformly decelerated with acceleration $a = \mu g$. The (horizontal component of the) translational velocity then follows the equation
$$
v(t) = v_0 - at
$$
Rotational motion
The dynamical friction acts with a torque of modulus $\mu mg R$ on the system (taking the center of the ball as a pole), then the angular acceleration is given by the rotation dynamic equation $I \alpha = \mu mgR$, which gives $\alpha = 2\mu g/(5R)$. The angular velocity $\omega$ as a function of time reads
$$
\omega(t) = \alpha t
$$
Pure rolling condition is obtained setting $v(t) = \omega(t) R$, which gives you the time needed to reach pure rolling, which is $t = v_0/(a+\alpha R)$. Now you can use this time to compute the distance with the kinematics law of the translation $D = v_0 t - (1/2)a t^2$.
I think you should account for friction in the energy balance of the problem, so $W=\Delta K_{trasl}+\Delta K_{rot}$, but then you would have two unknown variables: $D$ and the final velocity, so you should use kinematics as stated above in this case.
A: *

*since it's rolling without slipping the friction (static) won't affect the energy , so yes energy is conserved.


*here is a good demonstration https://www.youtube.com/watch?v=hxa6jAYA980 about similar case , after applying delta k= w ( transltaion )the forces you have are( only weight because again the friction you have is static
while applying delta k= w for rotation you have one torque which comes from friction
A: Hi there mister Radek Martinez! Good to see you've joined the game! And with a good question too!
Here's what I think:
The energy loss due to kinetic friction makes the ball move slower, thus the change in linear kinetic energy is equal to the loss in friction-induced loss of energy. Energy is left for linear and rotational velocity.
The force going hand in hand with kinetic friction is equal to the kinetic friction coefficient $\mu$ times the normal force acting on the mass, which is $F_n=Mg$, so:
$$F_{friction}=\mu Mg$$
This means the energy loss until the ball starts will be:
$$\int _0^D\mu Mgdx$$
So now we can write the equation for the conservation of energy:
$$\frac{1}{2}M {{{v_o}_{lin}}}^2=\int _0^D\mu Mgdx +\frac{1}{2}I{\omega}^2+ \frac{1}{2}M v_{lin}^2=\mu MgD+\frac{1}{2}I{\omega}^2+ \frac{1}{2}M v_{lin}^2,$$
where $v_{lin}$ with corresponding energy $\frac{1}{2}M v_{lin}^2$ is the linear motion of the ball when it's starting to move due to static friction and $\frac{1}{2}I{\omega}^2$is (a part) of the initial energy left for letting the ball rotate with static friction.
From this, we can easily extract $D$, the distance at which the ball starts to roll with static friction (all constants are known). I'll leave that for you to do.
I hope this answers the question (implicitly).
