[Experts only 3rd repost] What is terminal velocity of the puck?

Question

The puck slides along inclined plane. Initial velocity of the puck
is in horizontal direction (i.e. at right angle to plane inclination)
and has absolute value v0 = 1m/s.

http://alexandersemenov.tripod.com/puck/...

The angle of inclination is adjusted so, that coefficient
of kinetic friction is exactly equal to tan(a).


What is terminal velocity v1 of the puck?

http://alexandersemenov.tripod.com/puck/index.htm

Answer 1

Thank-you for the diagram!
I will assume that a is the angle subtended by the plane w/r/t the horizontal.

Since the coefficient of friction is equal to tan(a), gravity will only accelerate the puck while there is horizontal motion.

Since there is friction, the horizontal velocity will be dissipated, and only the velocity straight down the plane balanced with the kinetic friction will remain. Since the forces exactly balance, the puck will cease any acceleration in the direction straight down the plane.

look at a FBD when the puck is traveling only straight down the plane the forces acting on the puck are:

sin(a)*m*g-cos(a)*m*g*tan(a)=
m*acc

sin(a)-sin(a)=m*acc

So that proves that the forces balance

My intuition is telling me that the terminal velocity will be transposed from horizontal to parallel down slope, let's see if I can prove it mathematically.

The frictional force will always be sin(a)*m*g
let's call that f

The parallel component of gravity will always be
sin(a)*m*g, let's call that f also

During the transposition, the angle of the frictional f w/r/t the horizontal will vary, since friction is always tangential to motion.

When the puck is at angle(th) downslope w/r/t to horizontal the forces acting on the puck are
In the horizontal
cos(th)*f=m*acc h

note that for th=0,
acc h =f/m
and for th =90
acc h = 0

and the downslope will be
f-sin(th)*f=m*acc v
Note that for th= 0, acc v = f/m
for th = 90, acc v=0

When the puck moves an infinitesimal distance dl, the work done in the horizontal is:
-cos(th)*f*cos(th)*dl
-cos^2(th)*f*dl
note that
cos^2=1-sin^2

and the work done in the downslope is

-sin(th)*f*sin(th)*dl
-sin^2(th)*f*dl

note that these add to -dl*f

and the gain in kinetic energy due to vertical displacement is
dl*sin(th)*m*g*sin(a)

note that m*g*sin(a) is f
so dl*sin(th)*f

sum all of the energy together
sin(th)*dl*f-dl*f
this is the delta energy lost, since sin is
0 then there is energy lost as the puck transitions, so the terminal velocity will be the original kinetic energy of the system minus the integral of the energy . Since there are two variables, l, the distance traveled in the transposition, and (th), the instantaneous angle subtended, the integral must be simplified.

After thinking about this questin overnight and reviewing my calculus, I believe this can be solved using polar coordinates and partial differentiation. The motion path will follow an exponential relationship to Ehuler's constant, where the path is exponential in the horizontal and in the downslope. The force of friction will be related to this curve, and then the solution can be found using conservatin of energy.

Pretty complicated, but I don't see a simpler way.

j