Half pipe with friction in the middle

1. Mar 18, 2014

Dustinsfl

1. The problem statement, all variables and given/known data
Consider a half pipe of height L. The middle section, non sloping part, has a friction coefficient of $\mu_k = 0.1$ and frictionless every where else. The length of this section is L. How many times can the skateboarder go back and forth before he stops?

2. Relevant equations
$\sum\mathbf{F} = m\mathbf{a}$

3. The attempt at a solutionIn the friction section,
$$\sum F_x = v_x - F_f = v_x - .1N$$
since $F_f = \mu_k N$.
For the frictionless section, we would have (not sure about this part)
\begin{align}
F_y &= mg\cos(\theta)\\
F_x &= mg\sin(\theta)
\end{align}
Not sure how to determine how many times the skateboarder can go back and forth.

2. Mar 18, 2014

jackarms

I would think of it in terms of energy -- assuming he starts at the top of the half pipe, he has some potential energy, and then friction is going to take some away each time he passes across the middle section.

So you could say that:
$$\Delta E_{skateboarder} = -W_{friction}$$
And then it should be fairly straightforward.

3. Mar 18, 2014

Dustinsfl

If I use the conservation of energy, I have
$$mgL = mgh_2 + \frac{1}{2}mv_2^2$$
where I assumed the skateboarder started from rest.
Then what would I do?

Last edited: Mar 18, 2014
4. Mar 18, 2014

jackarms

Well, it won't exactly be conservation of energy -- think of it like this: if there was no friction, and the skateboarder started from rest at the top of the half pipe, then he could just go back and forth forever since his total energy (combined kinetic and potential) wouldn't change. But with friction, it will suck a bit of energy away each time he passes over the middle, so eventually his energy will become zero and so he will stop crossing over the middle. So think about how much energy friction is going to absorb each time, and then compare that to the energy he starts with.

5. Mar 18, 2014

Dustinsfl

So the energy is
$$mg(L-h_2) - \frac{1}{2}mv_2^2 = .1$$
and he started with $mgL$.
If this is correct, I am not sure what to do next.

Last edited: Mar 18, 2014
6. Mar 18, 2014

haruspex

How much energy does he lose each time he crosses the flat part?

7. Aug 13, 2015

Edward Folts

So it seems like the KE leaving the other side of the red line is equal to the KE(in) - .1KE(in)=.9KE(in). So KE(out)=.9KE(in)? Does this imply that there is an infinite amount of "back and fourths." It seems like it would be 0 (and thus stopping) at inifinity.

8. Aug 13, 2015

haruspex

That's not how friction works. It isn't taking 10% of the KE each traverse.
What equations do you know for the magnitude of kinetic friction and the work it does?

9. Aug 13, 2015

Edward Folts

Well the magnitude of the force of kinetic friction (not sure if I'm saying that correctly) would be (.1) * F(normal) = .1mg? and then the work done by it would be .1mLg?....If this is right I am going to face palm so hard...

10. Aug 13, 2015

Edward Folts

So with each pass, the skateboarder loses .1mgL worth of his/her kinetic energy. So there will be "x" passes until the KE is zero. So the initial KE energy going into the friction portion of the half pipe is mgL= (1/2)mv^2. And with each pass we subtrct .1mgL. So we have the equation (1/2)mv^2-.1mgL=0. And solving for x yields ((1/2)mv^2)/.1mgL = x = (mgL)/(.1mgL) = 10. So 10 passes?

11. Aug 13, 2015

Well done.