# Homework Help: Skateboarding physics

1. Nov 7, 2006

### Dorothy Weglend

I have a question about a skateboard problem. The athlete starts at the top of a half-pipe, in a crouched position, and goes to the bottom of the pipe. His center of mass moves along a circle with a radius of 6.3m.

Right after he reaches the bottom, he straightens out of his crouch, making his center of mass move along a circle of radius 5.85 m.

I am having trouble visualizing how the work done by the skateboarder straightening out of his crouch increases the potential energy which then goes into kinetic energy, increasing his velocity on the way up the other side of the half-pipe. It seems to me that this is perpendicular to the tangential motion of the skater, and so shouldn't have any effect on that part of the equation.

So if I need to calculate the velocity of the skater when he reaches the other top of the half pipe, I thought all I had to do was:

Kbottom = Utop + Ktop

And solve for the v in Ktop. But this results in an answer which is too small (according to the answer in the back of the book), which obviously used:

Kbottom + W_legs = Utop + Ktop

So I'm trying to understand why this would be right.

Thanks!
Dorothy

2. Nov 7, 2006

### turdferguson

Center of mass is simply the point that a large body's mass can be simplified to. When youve done problems in the past, youve assumed youre dealing with a point particle. But this is an example of the real world where CM can change. How high is the skateboarders CM in the beginning? How high is the CM at the end? It might make more sense if you calculate the velocity in the middle

3. Nov 7, 2006

### Dorothy Weglend

I have the velocity at the lowest point, is that what you mean by the middle?

Since the boarder doesn't come out of the crouch until after that point, it seems to me his height would be that of the original radius (6.3) not the new radius.

The problem is, he doesn't reach the top of the pipe without the extra work contributed by his legs. I'm having trouble understanding how work perpendicular to the motion contributes to that energy.

I am guessing it must be because the work is not truly perpendicular. He must lean at an angle, pushing against his skates, so this work goes into the kinetic energy of the skates.

Do you think this is right?

4. Nov 7, 2006

### turdferguson

I think its simpler than that. The CM is at its new lowest at the bottom of the pipe, set that to zero PE. At the top of the pipe, it has been raised 5.85m. In a pendulum or roller coaster, the velocity at the beginning height is the same as the velocity at the end height. But by raising the CM in this case, it loses less PE on the way up and therefore has more KE than it would otherwise. He would have more drag on the way up, but he would also lose less velocity

5. Nov 7, 2006

### Dorothy Weglend

The center of mass isn't raised at the top, since the athlete is parallel to the ground at that point (roughly). The vertical height is just that of the radius, which in this case I do believe is the original, crouching radius.

Basically, what you describe is what I am doing:

Kb + W = Ut + Kt

where 'b' is bottom, 't' is top, of course.

My main problem is trying to understand why a perpendicular 'W' should contribute anything.

Dorothy