Consider a right triangular lamina of areal density ρ, with one edge of length a along the x-axis and another edge of length b along the y-axis, as shown in the diagram.

(a) Find the center of mass (X,Y,Z) in this coordinate system.

(b) Find the components of the inertia tensor in this coordinate system.

(c)
For a = b, transform the inertia tensor to principal axes, giving the angle between the
principal axes and those shown in the diagram, and find the three moments of inertia in
the principal axes system.

(d) Use the parallel axis theorem to find
the principal moments of inertia about the center of mass.

Solution:

- Concepts:

The center of mass, the inertia tensor - Reasoning:

We chose a coordinate system. Then**R**_{CM}= (1/M)Σm_{i}**r**_{i}--> (1/M)∫**r**dm = (1/M)∫ρ(**r**)**r**dA.

I_{ij}= Σ_{k}m_{k}[Σ_{l}(x_{k})_{l}^{2}δ_{ij}- (x_{k})_{i}(x_{k})_{j}] --> ∫dm[Σ_{l}x_{l}^{2}δ_{ij}- x_{i}x_{j}] = ∫ρ(**r**)dA[Σ_{l}x_{l}^{2}δ_{ij}- x_{i}x_{j}] . - Details of the calculation:

(a) M**R**_{CM}=**i**∫dm x +**j**∫dm y

=**i**ρ∫_{0}^{b }dy ∫_{0}^{a-ay/b }dx x +**j**ρ∫_{0}^{a }dx ∫_{0}^{b-bx/a }dy y

= (ρab/6)(a**i**+ b**j**) = ⅓M(a**i**+ b**j**), since M = ρab/2.

X_{CM}= ⅓a, Y_{CM}= ⅓b, Z_{CM}= 0.

(b) I_{11}= ∫dm y^{2}= ρ∫_{0}^{a }dx ∫_{0}^{b-bx/a }dy y^{2}= ρab^{3}/12

I_{22}= ∫dm x^{2}= ρa^{3}b/12. I_{33}= ∫dm (x^{2}+ y^{2}) = (a^{2}+ b^{2}) ρab/12.

I_{12}= I_{21}= -∫dm xy = -ρ∫_{0}^{b }y dy ∫_{0}^{a-ay/b }dx = -ρa^{2}b^{2}/24. I_{13}= I_{23}= I_{31}= I_{32}= 0.

(c) For b = a we have

I_{11}= I_{22}= ρa^{4}/12 = Ma^{2}/6, I_{33 }= ρa^{4}/6 = Ma^{2}/3, I_{12}= I_{21}= -ρa^{4}/24 = -Ma^{2}/12, I_{13}= I_{23}= I_{31}= I_{32}= 0.

The inertia tensor in the given coordinate system is

I = (ρa

^{2}/12)a ^{2}-a ^{2}/20 -a ^{2}/2a ^{2}0 0 0 2a ^{2}.

To diagonalize this inertia tensor and find the principal moments of inertia I

_{x'}, I_{y'}, and I_{z'}, and the corresponding principal axes (x', y', z') we use L_{i}= Σ_{j}I_{ij}Ω_{j}= IΩ_{i}when the body is rotating about one of its principal axes. We write

a ^{2}- λ-a ^{2}/20 -a ^{2}/2a ^{2}- λ0 0 0 2a ^{2}- λΩ _{1}Ω _{2}Ω _{3}= 0.

Here I = ρa

^{2}λ/12. Since the matrix is block-diagonal we have immediately

I_{z'}= I_{33}= ρa^{4}/6 = Ma^{2}/3. The z'-axis is the z-axis.

To find I_{x'}and I_{y'}we solve (a^{2}- λ)^{2}- a^{4}/4 = 0, a^{2}- λ = ±a^{2}/2, λ_{x'}= a^{2}/2, λ_{y'}= 3a^{2}/2.

I_{x'}= ρa^{4}/24 = Ma^{2}/12, I_{y'}= ρa^{4}/8 = Ma^{2}/4.

Solving for the corresponding principal axes (the direction of**Ω**) we find

(a^{2}- λ_{x'})Ω_{1}- (a^{2}/2)Ω_{2}= 0, Ω_{1}- Ω_{2}= 0, Ω_{1}= Ω_{2}, 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 +45^{o}with the x-axis.

(a^{2}- λ_{y'})Ω_{1}- (a^{2}/2)Ω_{2}= 0, -Ω_{1}- Ω_{2}= 0, Ω_{1}= -Ω_{2}, 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 +135^{o}with the x-axis.

(d) Parallel axes theorem: I'_{ii}= I_{ii(CM) }+ Ma_{⊥}^{2}.

I_{z'CM}= I_{z'}- M(x_{CM}^{2}+ y_{CM}^{2}) = ρa^{4}/6 - (ρa^{2}/2)(2a^{2}/9) = ρa^{4}/18 = Ma^{2}/9.

I_{x'CM}= I_{x'}= ρa^{4}/24 = Ma^{2}/12/

I_{y'CM}= I_{y'}- Mx_{CM}^{2}+ y_{CM}^{2}) = ρa^{4}/8 - (ρa^{2}/2)(2a^{2}/9) = ρa^{4}/72 = Ma^{2}/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
trajectory.

For m = 1 kg, v = 500 m/s, λ = 45^{o} give a numerical answer.

M_{Earth} = 5.97*10^{24} kg, R_{Earth} = 6378 km.

Solution:

- Concepts:

Motion in an accelerating frame - Reasoning:

Assume an observer at mid latitudes λ = 90^{o}- θ.

The equations of motion in a rotating coordinate system contain fictitious forces.

md**v**/dt =**F**_{inertial}- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v**

= -m**g**+**N**+**F**_{ext}- m**Ω**× (**Ω**×**r**) - 2m**Ω**×**v.**-m**Ω**× (**Ω**×**r**) = centrifugal force.

(**Ω**×**r**) = ΩRsinθ**j**.**Ω**× (**Ω**×**r**) = -Ω^{2}Rsin^{2}θ**k**- Ω^{2}Rsinθcosθ**i**.

-2m**Ω**×**v**= Coriolis force.**Ω**×**v**= -Ωvcosθ**j.** - Details of the calculation:

We want d**v**_{tangential}/dt = 0. We need 0 =**F**_{ext}+ mΩ^{2}Rsinθcosθ**i**+ 2mΩvcosθ**j**.**F**_{ext}= -mΩ^{2}Rsinθcosθ**i**- 2mΩvcosθ**j**.

Plug in numbers.

Ω = 2π/(24*3600 s),**F**_{ext}= -1.7*10^{-2}N**i**- 5.1*10^{-2}N**j**.

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
**v**
and angular momentum **L** not quite parallel to the symmetry axis. Thus it
experiences precession.

(a) Calculate the principal moments of inertia of the spaceship about its
CM in terms of h,
r, and m_{1}.

(b) Show that the symmetry axis rotates in space about the fixed direction of the
angular momentum **L**.

Solution:

- Concepts:

Rigid body motion, principal moments of inertia - Reasoning:

We are asked to find the principal moments of inertia about the CM of a symmetrical top. It is easier to calculate the principal moments of inertia about the tip and then use the parallel axes theorem. -
Details of the calculation:

(a) Use cylindrical coordinates, with the origin at the origin of the primed system.

I_{1}' = ∫ρdV (y^{2}+ z^{2}), I_{2}' = ∫ρdV(x^{2}+ z^{2}), I_{3}' = ∫ρdV(x^{2}+ y^{2}),

I_{3}' = I_{3}= ρ∫_{0}^{h}dz∫_{0}^{ztanα }dr' 2πr'^{3}=^{ }2πρtan^{4}α ∫_{0}^{h}dz z^{4}/4 = 2πρtan^{4}α h^{5}/20.

ρ = m/V, V = ∫_{0}^{h}dz π(ztanα)^{2}= π tan^{2}α h^{3}/3.

With tanα = r/h we have ρ = 3m/(πhr^{2}) and I_{3}= 3mr^{2}/10.

From symmetry I_{1}' = I_{2}'.

2I_{1}' = ρ∫dV (x^{2}+ y^{2}+ 2z^{2}) = I_{3}+ 2ρ∫dV z^{2}.

I_{1}' = I_{2}' = I_{3}/2 + ρ∫dV z^{2}= I_{3}/2 + ρ∫_{0}^{h}dz πz^{4}tan^{2}α = ρπtan^{2}α h^{5}/5.

I_{1}= I_{2}= 3mr^{2}/20 + 3mh^{2}/5.

The CM is located a ρ = 0 and z_{CM}, where

z_{CM}= (ρ/m)∫dV z = (ρ/m)∫_{0}^{h}zdz π(ztanα)^{2}= (ρ/m)π tan^{2}α h^{4}/4 = 3h/4.

Using the parallel axes theorem we therefore have

I_{3}= I_{3}' = 3mr^{2}/10,

I_{1}= I_{1}' - m(3h/4)^{2}= (3/20)mr^{2}+ (3/80)mh^{2}= I_{2}.(b) L

_{1}= I_{1}ω_{1}, L_{2}= I_{1}ω_{2}, L_{3}= I_{3}ω_{3}.

We can chose the orientation of our space-fixed axes so that L_{1}= 0, ω_{1}= 0.

In the body-fixed coordinate system**L**,**ω**, and the z-axis lie in the same plane, but if I_{3}≠ I_{2},**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
±v**i**, collide elastically, head-on.
Assume that after the collision they have both reversed motion and are
still rolling.

(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?

Solution:

- Concepts:

Force and torque - Reasoning:

Each ball reverses the direction of its momentum and angular momentum. - Details of the calculation:

(a) Take the left ball. Let**k**point into the page and**j**point down.

Before the collision:**p**= mv**i**,**L**= Iω**k**= (Iv/R)**k**= (2mRv/5)**k**.

After the collision:**p**= -mv**i**,**L**= -(2mRv/5)**k**.

(**F**_{ball}+**F**_{table})Δt = -2mv**i**, (R**j**×**F**_{table})Δt = -(4mRv/5)**k**. (**F**_{ball}exerts no torque.)

**F**_{table}Δt = (4mv/5)**i**is the impulse the surface of the table exerts on each ball.

The impulse the table exerts on the left ball is (4mv/5)**i**, and the impulse the table exerts on the right ball is -(4mv/5)**i**.(b)

**F**_{ball}Δt = -(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 10^{o}?

(b) What energy (in Joules) is needed for the whole operation?

(Moment of Inertia of hollow spherical shell: [2M(R_{o}^{5 }-
R_{i}^{5 })]/[5(R_{o}^{3 }- R_{i}^{3})]
where R_{o} is outer radius and R_{i} is inner radius)

Solution:

- Concepts:

Conservation of angular momentum - Reasoning:

The system consists of the shell and the flywheel. No external torques act on the system, so the total angular momentum of the system is conserved. If a component of**L**is zero before the flywheel starts to rotate, then it is zero afterwards. - Details of the calculation:

(a) The flywheel receives an angular impulse, so the shell receives an angular impulse of equal magnitude and opposite direction.

L = Iω.

For the disk I = ½mr^{2}= 0.05 kgm^{2}. L = (0.05 kgm^{2})1000*2π/60s = 5.24 kgm^{2}/s.

For the shell I = [2M(R_{o}^{5 }- R_{i}^{5 })]/[5(R_{o}^{3 }- R_{i}^{3})] = 122664 kgm^{2}.

ω = dθ/dt = L/I = 4.27*10^{-5}/s.

t = (10*π/(180*4.27*10^{-5}))s = 4089 s ≈ 68 min.

(b) The energy needed is the final kinetic energy of the flywheel and the shell.

E = ½I_{w}ω_{w}^{2}+ ½I_{s}ω_{s}^{2}= 274.2 J.