Calculating Kinetic Energy After a Turn on a Horizontal Plane

[kaspis245]

Are you sure you copied the original problem text word-by word?

Absolutely yes. Notice that it's not mention that it's a ball. I think we were going in the right direction.

 

[ehild]

Absolutely yes. Notice that it's not mention that it's a ball. I think we were going in the right direction.

If you want to go ahead with the differential equation, see:
http://www.sosmath.com/diffeq/first/lineareq/lineareq.html
We have theta instead of x as the independent variable and z(theta) is the function instead of y(x),
$z' + 2μz =-\frac{2g}{R}\left(μ\cos(θ)+g\sin(θ)\right)$
So p(θ)=2μ and its integral is 2μθ. That is simple.
$u(θ)=e^{\int(p(θ)dθ}=e^{2μθ}$
Now the tricky part comes: Find the integral $S=\int{u(θ)g(θ)dθ}=-\frac{2g}{R}\int{e^{2μθ}\left(μ\cos(θ)+g\sin(θ)\right)dθ}$ Integrate by parts twice.
The solution is $z(θ)=\frac{S+C}{u(θ)}$
C is the integration constant,the initial value of z, (z at θ=0).

 

[mfb]

As the radius of curvature is not given, the intended answer must be independent of it. This is not true in general - even without friction the loss to gravity will depend on R. It is true if we let R be very small and treat the mass as (non-rolling) point, however. That leads to a unique solution that depends on the given quantities only. Not sure if you can get away without integration, but it certainly makes the equations easier.

 

[ehild]

As the radius of curvature is not given, the intended answer must be independent of it. This is not true in general - even without friction the loss to gravity will depend on R. It is true if we let R be very small and treat the mass as (non-rolling) point, however. That leads to a unique solution that depends on the given quantities only. Not sure if you can get away without integration, but it certainly makes the equations easier.

So you suggest to solve the problem for the limiting case R-->0, but still assuming smooth transition between the two planes.
In case of very small radius, the transition from one plane to the other happens in a very short time. It is collision, in principle. The normal force is impulsive, and gravity can be ignored with respect to it. Friction is also impulsive, as it it proportional to the normal force. The change of kinetic energy is equal to the work of all forces. The work of the normal force is zero, The work of gravity can be ignored, so it is the work of friction alone to be counted with.
The change of KE is equal to the work of friction $dE= -μN(Rdθ)$ and N is equal to the centripetal force if mg is ignored : $N=\frac{mv^2}{R}$, so
$dE= -μ\frac{mv^2}{R}(Rdθ) = -μmv^2dθ$.
$E=1/2 mv^2$, so we can write a DE for the kinetic energy:
$\frac{dE}{dθ} = -2μE$.
It is very easy to solve and get the kinetic energy at the final angle α.
@https://www.physicsforums.com/members/kaspis245.534046/

 

[haruspex]

It is collision, in principle.

Treating it as an inelastic collision would mean that the component of velocity normal to the ramp would be lost in the impact, independently of any friction. Your analysis above does not do that, but I agree with mfb that that analysis is exactly what is wanted in answering the question.

 

[ehild]

I meant considering it a collision in the sense that it happens in a very short time: The normal force is impulsive and the effect of gravity can be ignored.
My previous approach which assumed a certain R leads to the same result at the limit R-->0. But I think, the method based on work-energy theorem outlined in Post ##34 might be more familiar to the OP.