Ball rolling down an incline and dynamics in general I recently performed an experiment for a class which involved measuring the projectile motion of a ball after it had rolled down a ramp and exited on a significantly smaller ramp. When interpreting the measurements taken (which include the time taken for the ball to leave the ramp, diameter of the ball along with a video of the flight path), I made the mistake of applying the motion of an object sliding down a plane instead of that of a ball rolling down a plane- which I have not covered and only have limited mathematical understanding of. 
I have found a number of formulas online including Angular acceleration, Moment of Inertia and Rolling friction but I am either not using them correctly, I am missing something or I am completely overthinking the process. The ultimate goal of this is to find the velocity of the ball as it leaves the ramp, am I just overthinking this? 
 A: I would assume that the friction does not dissipate energy but only causes the ball to roll without slipping. Then, the potential energy
$$E_p = m g h$$  will be converted to kinetic energy
$$E_{pot} = \frac{1}{2} m v^2$$
and to angular momentum. The energy due to the angular momentum can be calculated via the moment of inertia, which, for a solid ball is 
$$I= \frac{2}{5} m r^2$$
with corresponding energy
$$E_{rot}= \frac{1}{2} I \omega^2.$$
With the condition for the rolling and not slipping
$$\omega=\frac{v}{r}$$
you should be able to solve the problem
$$E_{pot}=E_{rot}+E_{kin}.$$
Good luck!
A: 
FORCE/TORQUE METHOD:

$mg\sin(\theta)-f=ma$
$fR=I\alpha$
$a_c=R\alpha$

$\implies a_c=\dfrac{g\sin(\theta)}{(1+k^2 /R^2)}$ and $f=\dfrac{mg\sin(\theta)}{(1+R^2/k^2)} $ 
$v_{final}^2=0+2(a_c)(\dfrac{h}{\sin(\theta)})$
Solve to get $v_{final}$
MECHANICAL ENERGY CONSERVATION:

$0+mgh=\dfrac 12 mv_{final} ^2 +\dfrac 12 I_{c}\omega_{final}^2$
$v_{final}=R\omega_{final}$

$W_{friction}=0$ since point of contact of the spherical body with inlined plane is always instantaneously at rest. 
Solve to get $v_{final}$
WORK-ENERGY THEOREM SEPARATELY IN TRANSLATIONAL MOTION AND ROTATIONAL MOTION:

$-f(\dfrac{h}{\sin(\theta)})+mgh=\dfrac{1}{2}mv_{final}^2$
$(fr)(\dfrac{h/\sin(\theta)}{r})=\dfrac{I_{c}w_{final}^2}{2}$ [Toque *
  Angular displacement = Final Rotational KE]

Solve to get $v_{final}$
ANGULAR IMPULSE MOMENTUM THEOREM METHOD:
$L=L_{translational}+L_{rotational}$
By applying the impulse-momentum theorem at any random point at time $t$ and $t+dt$:

$( \dfrac{I_{c}}{r_{final}} + mr )(v)(- \hat k) + mgr\sin(\theta)dt
 (-\hat k)=( \dfrac{I_{c}}{r_{final}} + mr )(v + dv)(- \hat k)$

Find $\dfrac{dv}{dt}$ from above equation which equals $a_{c}$ and then use it to find final velocity at bottom as shown previously.
You can use linear impulse momentum theorem also like:
$mv +mg\sin(\theta) dt -fdt = m(v+dv)$ to solve the problem.Just reuse the value of $f$ we obtained in first method! 
Hope this helps!
