Often when a speed is required, energy is an efficient way to obtain the answer. In this particular problem, angular momentum is a much faster route, but it is instructive to consider the form of the energy equations.
As the ball rolls along the alley, its height is not changing and so the gravitational potential energy is fixed (and so can be set to zero). The normal force is perpendicular to the ball's linear movement and also has no [moment arm] about the ball's center of mass, and so does no work that would affect the ball's linear velocity or its angular velocity. Friction, however, is anti-parallel to the ball's displacement and so it does negative work on the ball's translational kinetic energy and it has a non-zero lever arm so it does work that acts to increase the ball's angular speed. The appropriate form of the mechanical energy equation is:
{latex}\begin{large} \[ W_{nc}+E_{i} = -F_{f}x+F_{f}R\theta + \frac{1}{2} mv_{i}^{2} + \frac{1}{2}I\omega_{i}^{2} = E_{f}=\frac{1}{2} mv_{f}^{2} + \frac{1}{2}I\omega_{f}^{2}\]\end{large}{latex}
{note}Note that friction contributes two forms of non-conservative work, one from the linear motion and one from the rotational motion of the ball.{note}
{warning}It is very important to recognize that mechanical energy is _not_ conserved in this problem. The ball is _not_ rolling without slipping, it is sliding and rotating. Thus, the friction is _not_ static and so it does dissipate some mechanical energy into heat. For examples where pure rolling without slipping gives conservation of mechanical energy even in the presence of friction see, e.g.:
{contentbylabel:rolling_without_slipping,energy_conservation|operator=AND|showSpace=false|showLabels=false}{warning}
Once again, we know that ω{~}f~ R = _v{_}{~}f~. We *do not*, however, know any relationship between _x_ and θ at this time. The ball _ends up_ rolling without slipping, giving the relationship between the final speeds, but it has certainly been slipping in the interim. Thus, the angular distance covered will not be directly related to the linear distance. To solve this problem, we must assume that the acceleration and angular acceleration are constant. In that case, we know from kinematics that:
{latex}\begin{large}\[ x = \frac{v_{f} + v_{i}}{2} t \] \[ \theta = \frac{\omega_{f} + \omega_{i}}{2} t\]\end{large}{latex}
We also know, as was used in Method 2, above, that:
{latex}\begin{large}\[a_{x} = - \frac{F_{f}}{m} = \frac{(v_{f} - v_{i})}{t}\]\end{large}{latex}
Substituting appropriately into the energy equation and using the rolling without slipping condition then gives the same answer as was obtained via methods 1 and 2.
{info}It is instructive to note that the (negative) work done by friction on the linear kinetic energy is:
\\
{latex}\begin{large}\[ -F_{f} x = -F_{f}\frac{v_{f}+v_{i}}{2} t \]\end{large}{latex}
while the positive work done in increasing the angular kinetic energy is:
{latex}\begin{large}\[ F_{f}R\theta = F_{f} \frac{v_{f}}{2} t \] \end{large}{latex}
Clearly, the negative work is larger in magnitude, and we can see that the net energy dissipated to heat will be:
{latex}\begin{large} \[ |W_{\rm dissipated}| = F_{f} \frac{v_{i}}{2} t \] \end{large}{latex}
where _t_ is the amount of time elapsed from the instant the ball is released until the instant it begins to roll without slipping.
{info}
| 
