The key to reconciling the lost energy to the dissipation that occurs
at the slipping surface is to recognize that the rate at which energy
is dissipated is the frictional force times the *relative slipping
velocity* of those surfaces. Since that relative velocity decreases
linearly with time to zero at the end of the slipping period, so does
the rate of energy dissipation. Thus, the total energy dissipated is
just the total time of slipping times *half* the initial dissipation
rate, where the initial dissipation rate is simply the frictional
force times the initial tangential velocity of the object.
For those one or two brave souls who care to sweat the details, here
is the full analysis:
Assume that the wheel starts out spinning around a horizontal axis,
at rest translationally, in contact with the floor, and in vertical
equilibrium.
Work in units where R = the radius of the object = 1, m = mass of
object = 1, and wo = initial angular velocity = 1.
The problem then depends on only the following three dimensionless parameters:
mu = coef of friction, g = grav field strength/(wo^2*R), r = radius
of gyration/R
It is not hard to show that
Angular speed/wo w(t) = 1 - mu*g*t/r^2
Translational speed/(wo*R) v(t) = mu*g*t
Frictional force/(m*wo^2*R) f = mu*g
Duration of slipping*wo T = r^2/[mu*g*(1+r^2)]
Thus, the rate at which energy is dissipated is
P(t) = f*(w*R-v) = mu*g[1 - mu*g*t*(1+r^2)/r^2]
This shows that the energy dissipation rate decreases linearly with
time from an initial value of mu*g to zero at the end of the slipping
interval. [Note that the properly dimensioned initial dissipation
rate is mu*m*g*wo*R.]
The dissipation rate is now easily integrated to get a result that is
completely consistent with that obtained using conservation of
angular momentum:
Total energy dissipated = (1/2)*r^2/(1+r^2)
or in properly dimensioned form, (1/2)*m*wo^2*R^2*r^2/(1+r^2)