# Energy Conservation in Rolling without Slipping Scenario

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.

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