I = (ρa2/12)
To diagonalize this inertia tensor and find the principal moments of inertia Ix',
Iy', and Iz', and the corresponding principal axes (x',
y', z') we use Li = ΣjIijΩj = IΩi
when the body is rotating about one of its principal axes. We write
|a2 - λ||-a2/2||0|
|-a2/2||a2 - λ||0|
|0||0||2a2 - λ|
Here I = ρa2λ/12. Since the matrix is block-diagonal we have
Iz' = I33 = ρa4/6 = Ma2/3. The z'-axis is the z-axis.
To find Ix' and Iy' we solve (a2 - λ)2 - a4/4 = 0, a2 - λ = ±a2/2, λx' = a2/2, λy' = 3a2/2.
Ix' = ρa4/24 = Ma2/12, Iy' = ρa4/8 = Ma2/4.
Solving for the corresponding principal axes (the direction of Ω) we find
(a2 - λx')Ω1 - (a2/2)Ω2 = 0, Ω1 - Ω2 = 0, Ω1 = Ω2, The components along of Ω along the x and y-axis are equal to each other, the principal axis points in the (i + j) direction, it makes an an angle of +45o with the x-axis.
(a2 - λy')Ω1 - (a2/2)Ω2 = 0, -Ω1 - Ω2 = 0, Ω1 = -Ω2, The components along of Ω along the x and y-axis are equal in magnitude and have opposite signs. The principal axis points in the (-i + j) direction, it makes an an angle of +135o with the x-axis.
(d) Parallel axes theorem: I'ii = Iii(CM) + Ma⊥2.
Iz'CM = Iz' - M(xCM2 + yCM2) = ρa4/6 - (ρa2/2)(2a2/9) = ρa4/18 = Ma2/9.
Ix'CM = Ix' = ρa4/24 = Ma2/12/
Iy'CM = Iy' - MxCM2 + yCM2) = ρa4/8 - (ρa2/2)(2a2/9) = ρa4/72 = Ma2/36.
Assume a perfectly spherical Earth of radius R with a
frictionless surface. On the surface of this Earth an object with mass m is
moving with constant speed v towards the north pole. When the object is at
latitude λ, find the external force required to keep it moving on that
For m = 1 kg, v = 500 m/s, λ = 45o give a numerical answer.
MEarth = 5.97*1024 kg, REarth = 6378 km.
A rigid, symmetrical spaceship is shaped in the form of a
cone with a uniform density. The height of the cone is h, the radius of the base is
r, and the total mass is m. Being suspended in outer space
without any external forces acting on it, the space ship has a center of mass velocity
and angular momentum L not quite parallel to the symmetry axis. Thus it
(a) Calculate the principal moments of inertia of the spaceship about its CM in terms of h, r, and m1.
(b) Show that the symmetry axis rotates in space about the fixed direction of the angular momentum L.
(b) L1 = I1ω1, L2 = I1ω2,
L3 = I3ω3.
We can chose the orientation of our space-fixed axes so that L1 = 0, ω1 = 0.
In the body-fixed coordinate system L, ω, and the z-axis lie in the same plane, but if I3 ≠ I2, L and ω are not parallel to each other.
The direction of L is fixed in space.
Consider a point P on the z-axis of the top, displaced by r from the CM.
With respect to space-fixed coordinates, the velocity of point P is v = ω × r.
At every instant the velocity of points on the z-axis is perpendicular to the plane containing L, ω, and the z-axis, so the z-axis must rotate about L.
Two identical billiard balls of radius R and mass M, rolling with CM velocities
±vi, collide elastically, head-on.
Assume that after the collision they have both reversed motion and are
(a) Find the impulse which the surface of the table must exert on each ball during its reversal of motion.
(b) What impulse is exerted by one ball on the other?
FballΔt = -(2mv + 4mv/5)i = -(14mv/5)i is the impulse the right ball exerts on the left ball.
The impulse the left ball exerts on the right ball is (14mv/5)i.
A stationary space station can be approximated as a hollow spherical shell of
mass 6 tons (6000 kg) and inner and outer radii of 5 m and 6 m. To change its
orientation, a uniform fly wheel of radius 10 cm and mass 10 kg located at the
center of the station is spun quickly from rest to 1000 rpm.
(a) How long (in minutes) will it take the station to rotate by 10o?
(b) What energy (in Joules) is needed for the whole operation?
(Moment of Inertia of hollow spherical shell: [2M(Ro5 - Ri5 )]/[5(Ro3 - Ri3)] where Ro is outer radius and Ri is inner radius)