Physics Friday 94

A classic problem: A skier starts at rest at the top of a large, hemispherical hill with radius (and thus height), R. If friction is negligible, at what height h above the base of the hill will she become airborne? What is her velocity, both magnitude and angle below the horizontal, at this point?

At any point before the skier becomes airborne, she is acted on by two forces; gravity, and the normal force exerted by the hill. If she is at a point on the hill which is at an angle θ from the top of the hill (as measured at the center of the hill), and moving at speed v, then the normal force N will be at an angle θ from vertical. As she is undergoing non-uniform circular motion, the net force will have both radial and tangential components; however, the radial component must still be centripetal force  inward, where m is her mass.

Now, the (inward) radial component of the force is ; equating, we see
.
Now, the skier becomes airborne when the normal force the hill exerts on her becomes zero. The above equation tells us that this occurs when the angle θ and speed v are such that . But, we note from trigonometry that , so that we have
.

Now, we could use the tangential component  of the force to find the tangential acceleration  as a function of angle, and thus position; this gives a second-order differential equation. However, there is a much easier method to find the velocity as a function of position. Since there is no friction, we can use conservation of energy, with the only potential energy being gravitational. At the top of the hill, the skier has only potential energy . At height h, the potential energy is the smaller , and the kinetic energy is just . Conservation of energy tells us, then, that
; we can cancel out the skier’s mass, which appears in all terms, and then solve for v2 to get
.
Plugging this into our relation for the point where the skier goes airborne, we get
.

Thus, we have the height where the skier becomes airborne, which is independent of the skier’s mass (and the acceleration of gravity g). This corresponds to an angle θ of
. As her velocity is still tangential at the point where she becomes airborne, this is also the angle below horizontal of her velocity at that point. That velocity has magnitude
.

[Update 7:15 – Fixed an error in description of tangential acceleration method].