The tilted hemisphere.
A solid hemisphere of radius R rests on a surface rough enough so it doesn't slip. The ramp tilt is φ.
1. Assuming that friction is sufficient that the hemisphere doesn't slip. What is the maximum angle of the tilted plane that can allow the hemisphere to rest on the plane in static equilibrium?
2. Resting at smaller angles of tilt, if the hemisphere is tweaked a bit it will vibrate for a brief while. What is the period of this small angle motion? How does this depend on the mass of the hemisphere or the angle of the plane?
3. A solid hemisphere rests on a level, flat surface, with its flat side down. It is given a forceful spin. It rises to spin on its edge a while, then settles down as it loses energy. In its final resting position is its flat face up, or is its flat face down? Before you answer, note that the hemisphere held at rest with flat side vertical will fall so it ends up at rest with flat side up. The student says, "Of course, the center of mass is not at the hemisphere's center, but is toward the curved side, so that side will fall." Does this appply to the case of the hemisphere spinning on its edge.
In all cases, show your reasoning and mathematical analysis.
Hints.1. The center of mass of a solid hemisphere of radius R is 3R/8 from the flat face. You can look this up, or find its derivation many places on the web.
At equilibrium on the tilted plane, the hemisphere's axis tilts from the vertical at the same angle the plane tilts to the horizontal.
2. At smaller inclination angles of the plane, the hemisphere can be in static equilibrium, its axis tilted at the same angle as the plane's inclination angle, At zero inclination the hemisphere can oscillate as a physical pendulum. So it also does so on the tilted plane, but does the tilt change the oscillation period?
We can be confident that the period will not depend on the hemisphere's mass. But if you don't know why, perhaps you should start with the more fundamental question, why is the period of a simple pendulum independent of mass? Why do bodies of different mass in a vacuum fall equal distances in equal times?
3. The hemisphere always lands flat side down, contrary to expectations. If you try this, the spin is so rapid you see only a blur. A high speed video camera should be able to capture this in slow motion. Does any reader have the equipment to do this? But this would show only what happens, not why it happens. It cannot directly reveal the forces acting on the hemisphere. And, unless the hemisphere has a pattern painted on it, it won't reveal whether the spin of the hemisphere about its own axis remains in the same direction, or whether it is rolling or slipping on the flat surface.
As the hemisphere begins to wobble, curved side down, friction due to the floor provides a torque that does work on the hemisphere, causing its center of mass to rise. Once this motion begins, the outcome is already inevitable This torque adds a momentum component along the axis of the hemsphere, causing the center of mass of the half-sphere to rise in height above the floor. At the highest point of the center of mass, this momentum begins to decrease, but has already caused the hemisphere to "cross over" to spin with its curved face downward until it settles on to the floor. Too simplistic? Of course the devil is in the details.
1. At small angles of stable equilibrium the center of mass, G, is directly above the point of contact, P, with the tilted plane. Since there are only two forces acting on the hemisphere, they must be colinear for two reasons (1) this ensures zero torque on the hemisphere, and (2) The horontal component of the net force on the hemisphere must be zero, since at equilibrium the body isn't moving.
As the pane's inclination angle increases, the hemisphere rolls until the situtation shown in Fig. 2 is reached. If it rolled further, the point G cannot be above P, and the hemisphere falls to the right. The points P, G, and C form a right triangle.
The maximum tilt angle is threfore φ = asin((8/3)sin(α)) = 22.024313...°.
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