Is the cone’s own center line, another, x 2, Next, we first define a new set of axes with origin O: So the angular velocity Ω = V a sin α = θ ˙ cot α. Now visualize the rolling cone turning around the So contributes translational kinetic energy 1 2 M V 2 = 1 2 M θ ˙ 2 a 2 cos 2 α. The center of mass moves at velocity V = θ ˙ a cos α, Moves along a circle at height a sin α above the plane, this circle being centered on Of mass, which is a distance a from the vertex, and on the central line, Have semi-vertical angle α (meaning this is the angle between O A and the central axis of the cone) the center Therefore, the angular velocity vector Ω → points along O A.
This means that, at that moment, the cone is rotating about the stationary line O A. Momentarily at rest when it’s in contact with the plane. The important point is that this line of contact, regarded as part of the rolling cone, is The momentary line of contact with the plane is O A,Īt an angle θ in the horizontal plane from the X axis. The cone rolls without slipping on the horizontal X Y plane. Analyzing Rolling Motion Kinetic Energy of a Cone Rolling on a Plane The moment of inertia about the axis x ′ 1 through the vertex, perpendicular to theĬentral axis, can be calculated using the stack-of-discs parallel axisĪpproach, the discs having mass π ρ R z h 2 d z, The moment of inertia about the central axis of the cone is (taking density ρ ) that of a stack of discs each having mass m d z = π r 2 ρ d z = π R z h 2 ρ d z and moment of inertia I d z = 1 2 m d z r 2:
The center of mass is distance a from the vertex, whereĪ V = a ⋅ 1 3 π R 2 h = ∫ 0 h z d V = ∫ 0 h π z R z h 2 d z = 1 4 π R 2 h 2, a = 3 4 h. V = ∫ 0 h π r 2 d z = ∫ 0 h π R z h 2 d z = 1 3 π R 2 h. Motion of inertial about their common axis (shown).įollowing Landau, we take height h and base radius R and semivertical angle α so that R = h tan α.Īs a preliminary, the volume of the cone is Revolution and a sphere of the same mass and radius clearly have the same įor a sphere, a stack of discs of varying radii, I = ∫ − a a d z 1 2 ρ π a 2 − z 2 2 = 8 15 ρ π a 5 = 2 5 M a 2. Regarded as a stack of discs, of radius a, A solidĬylinder about a line through its center perpendicular to its main axis can be
This is also correct for a cylinder (think ofĪ disc about a line through its center in its plane must be 1 4 M a 2 from the perpendicular axis theorem. Radius a and surface density σ has I = ∫ 0 a r 2 ⋅ σ ⋅ 2 π r d r = 1 2 π a 4 σ = 1 2 M a 2. Line perpendicular to the plane through the center is twice this -thatįormula will then give the moment of inertia of a cube, about any axis through Line in the plane through the center, from the symmetry, and the moment about a (It’s just a row of rods.) in fact, the moment is the same about any Ībout an axis in its plane, through the center, perpendicular to a side: I = 1 12 m ℓ 2. We have spherical degeneracy, meaning all three principal axes have the same Here we have four obvious principal axes: only possible if What about a symmetrical three dimensional molecule?
The moment of inertial about either of theseĪxes will be one-half that about the perpendicular-to-the-plane axis. Obviously, one principal axis is through the centroid,Įstablished that any axis of symmetry is a principal axis, so there areĮvidently three principal axes in the plane, one along each bond! The only interpretation is that there is aĭegeneracy: there are two equal-value principal axes in the plane, and any two Lowest angular momentum state E = L 2 / 2 I = ℏ 2 / 2 I , is lower for molecules with bigger moments of This was not the case for heavier diatomic gases, since the energy of the Was so low that a lot of energy was needed to excite the first quantized The resolution was that the moment of inertia Though they constantly collided with each other. Temperatures, apparently these diatomic molecules didn’t spin around, even Which gave an excellent account of the specific heats of almost all gases, In the nineteenth century, the mystery was that equipartition of energy, Two point masses m a fixed distance a apart gives I = 1 2 m a 2. The moment of inertia of the hydrogen molecule wasįind: the nuclei (protons) have 99.95% of the mass, so a classical picture of Michael Fowler Examples of Moments of Inertia Molecules
Īlso note that unlike the second moment of area, the product of inertia may take negative values.Previous home next PDF 24. Principal axes Reference Table Area Moments of Inertia