Airtime for halfpipe skateboarder

[ec-physics]

Homework Statement

A skateboarder drops in from one end of a halfpipe, and travels to the top of the other end of the halfpipe. He needs two seconds to perform a twist.
Given the height of a halfpipe, and the mass of the skateboarder, what would be the skateboarder's maximum air time, and would he be able to perform his twist?

Halfpipe height h: 3.66m
Skateboarder mass m: 39.5 kg

Relevant Equations

PE=mgh
KE=1/2mv^2
vf = vi + at

I've calculated the potential energy at the top of the halfpipe, before the boarder drops in:
PE = 39.5 kg * 9.8 m/s^2 * 3.66 m = 1416 J

Since the boarder would have no potential energy and all kinetic energy at the bottom of the halfpipe,
KE = 1/2mv^2 = 1416 J
1/2 (39.5 kg) (v^2) = 1416 J
So his velocity at the bottom of the halfpipe is 8.46 m/s^2.

I'm getting stuck with translating that into the second half of the halfpipe journey, when he begins traveling up, catches air, and comes back down. Where do I go from here? I feel like I'm simplifying this into just kinetic and potential energy when there are more factors at play, like momentum.

 

[jbriggs444]

You should keep symbolic quantities symbolic as long as possible. This makes it easier to do the algebra. If you had done this you could have noticed that the skater's mass is irrelevant. You multiplied and divided by 39.5 for no reason.

As the skater comes up the other side of the half pipe, his kinetic energy is converted to potential. How much kinetic energy should he have left when he returns to the original starting height?

Edit: It seems that the problem statement is missing information. I can see a way to solve if we make an assumption about the means by which the skater gains kinetic energy (he rises from a crouch at the very bottom of the half-pipe), an assumption about the half-pipe profile (semi-circle, vertical ends and radius 3.66 meters), an assumption about how many cycles he has to gain speed (one pass), an assumption about his starting velocity (starting from rest) and an assumption about how deeply he is able to crouch (say 0.5 meters).

 

[ec-physics]

You should keep symbolic quantities symbolic as long as possible. This makes it easier to do the algebra. If you had done this you could have noticed that the skater's mass is irrelevant. You multiplied and divided by 39.5 for no reason.

As the skater comes up the other side of the half pipe, his kinetic energy is converted to potential. How much kinetic energy should he have left when he returns to the original starting height?

Since he's back to the same height, at TE=PE+KE, he should have no kinetic energy and all potential, right?
So in theory he wouldn't ever get above the 3.66m halfpipe?

This is why I'm pretty sure I'm missing something, or pursuing the wrong avenue.

 

[jbriggs444]

Since he's back to the same height, at TE=PE+KE, he should have no kinetic energy and all potential, right?

So in theory he wouldn't ever get above the 3.66m halfpipe?

Right.

This is why I'm pretty sure I'm missing something, or pursuing the wrong avenue.

See my edited response above for an avenue by which a solution might be pursued.

 

[ec-physics]

Right.

See my edited response above for an avenue by which a solution might be pursued.

So if he does change his height, that would change the radius from the center of the halfpipe (assuming it's a semi-circle), which would change... angular velocity? I've looked at this link briefly: https://www.real-world-physics-problems.com/physics-of-skateboarding.html
Think it could point me down the right road to finding out air time?

 

[jbriggs444]

So if he does change his height, that would change the radius from the center of the halfpipe (assuming it's a semi-circle), which would change... angular velocity? I've looked at this link briefly: https://www.real-world-physics-problems.com/physics-of-skateboarding.html
Think it could point me down the right road to finding out air time?

Right. That is one way to attack the question of how much kinetic energy is gained. Rising from the crouch changes his "moment of inertia". Angular momentum is conserved, so angular velocity must change inversely.

That gives you a starting point with which to compute the resulting velocity and energy.

A different approach which should yield the same final result would consider the work done by the skater's legs lifting himself against the centrifugal force field.

 

[ec-physics]

Right. That is one way to attack the question of how much kinetic energy is gained. Rising from the crouch changes his "moment of inertia". Angular momentum is conserved, so angular velocity must change inversely.

That gives you a starting point with which to compute the resulting velocity and energy.

A different approach which should yield the same final result would consider the work done by the skater's legs lifting himself against the centrifugal force field.

Question: how would I go about considering the amount of work done by the skater's legs? Would that be just an estimate based on how much a person can push with their legs?

 

[jbriggs444]

Question: how would I go about considering the amount of work done by the skater's legs? Would that be just an estimate based on how much a person can push with their legs?

You know the velocity at the bottom of the curve.
That tells you the centripetal acceleration at the bottom of the curve (and along trajectories with narrower radii)
If the person has the strength to rise out of the crouch, that tells you how much force he needs to push with. A difficulty is that the tangential velocity will change due to the skater's efforts. So the centrifugal force field does not have a simple potential function. You probably do not want to go this way.

The angular momentum approach is much simpler. You can get the resulting velocity change almost immediately.

 

[kuruman]

Simple calculation: In 2 seconds an object released from rest drops distance
$$d=\frac{1}{2}gt^2=\frac{1}{2}\times 9.8~ \rm{(m/s^2)}\times 2^2~\rm{(s^2)}=19.6 \rm~{m}$$
This means that he needs to rise at least up to that height above the bottom of the half pipe. I am not a skateboarder but I think 2 seconds of air time is unrealistic.

 

[jbriggs444]

Simple calculation: In 2 seconds an object released from rest drops distance
$$d=\frac{1}{2}gt^2=\frac{1}{2}\times 9.8~ \rm{(m/s^2)}\times 2^2~\rm{(s^2)}=19.6 \rm~{m}$$
This means that he needs to rise at least up to that height above the bottom of the half pipe. I am not a skateboarder but I think 2 seconds of air time is unrealistic.

You can cut that vertical height by a factor of four if you realize that two seconds hang time is one second up and one second back down. It's still pretty high, but not outlandishly so.