# 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