Help needed with physics question

Question

A 58.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.70 m. What speed does he need at the bottom?

Any help would be appreciated, thank you.

Answer 1

A classic conservation of energy problem.

We need to find the total energy of the skateboarder at two points; the bottom of the ramp, and the upper edge of the "quarter pipe". Let's put the origin of our coordinate system on the ground (we only need the vertical direction for this problem, so you may put it where ever you like horizontally).

At the bottom of the ramp, the skateboarder has no height, and therefore no potential energy. He does, however, have some velocity (which we do not yet know) which we might as well label as v. Together with his given mass, we can find his total energy at the bottom is:

T_bot=U_bot+V_bot=(1/2)mv^2

We do the same at the top. Since he wants to "just make it" to the upper edge, he must reach it just as his velocity reaches 0 before falling back down. Therefor, his kinetic energy is 0. On the other hand, he has some distance above the ground, so he will have potential energy. If you sketch the quarter pipe, you should be able to see that that height is just the radius, r. So

T_top=U_top+V_top=mgr

Because energy is conserved, T_bot=T_top, so we can solve for the unknown v

(1/2)mv^2=mgr

v^2=2gr

v=sqrt(2gr)

Using g=9.81 m s-2, we get v=7.28 m s-1