__Center of mass__

A uniform carpenter's square has the shape of an L, as shown in the figure. Locate the center of mass relative to the origin of the coordinate system.

Solution:

- Concepts:

The center of mass - Reasoning:

We are asked to find the CM of an object.

We can think of the system as being made up of two subsystems, as shown in the figure.The CM of the left subsystem lies at x

_{CM }= 2 cm, y_{CM }= 9 cm. The CM of the right subsystem lies at x_{CM }= 8 cm, y_{CM }= 2 cm. If the mass of a 1 cm by 1 cm square is 1 unit, then the mass of the left subsystem is 72 units and the mass of the right subsystem is 32 units. We find the CM of the system by treating each subsystem as a separate particle, with all its mass concentrated at its center of mass. - Details of the calculation:

x_{CM }= (72 units 2 cm + 32 units 8 cm)/(104 units) = 3.85 cm,

y_{CM }= (72 units 9 cm + 32 units 2 cm)/(104units) = 6.85 cm.

The CM of the system lies outside of the system. For irregular-shaped objects it is quite common for the CM to lie outside the system. This special point outside the system responds to external forces as if the total mass of the system were concentrated there.

__Moment of inertia__

The four particles in the figure below are connected by rigid rods.

The origin is at the center of the rectangle. If the system rotates in
the x-y plane about the z-axis with an angular speed of 6 rad/s, calculate

(a) the moment of inertia of the system about the z-axis and

(b) the
rotational energy of the system.

Solution:

- Concepts:

The moment of inertia about an axis, the rotational kinetic energy - Reasoning:

The moment of inertia is I = ∑m_{i}r_{i}^{2}. Here r_{i}is the perpendicular distance of particle i from the z-axis.

the rotational kinetic energy, T_{rot}= ½Iω^{2} - Details of the calculation:

Each particle is a distance r = (9 + 4)^{½ }m = √(13) m from the axis of rotation.

I = (3 kg + 2 kg + 4 kg + 2 kg)(13 m^{2}) = 1 43 kgm^{2}.

(b) The rotational kinetic energy is T_{rot}= ½Iω^{2 }= 71.5*36/s^{2 }= 2574 J.

Three particles are connected by rigid rods of negligible mass lying along the y-axis as shown.

If the system rotates about the x-axis with angular speed of 2 rad/s, find

(a) the moment of inertia about the x-axis and the total rotational kinetic
energy evaluated from ½Iω^{2}, and

(b) the linear speed of each particle and the total kinetic energy evaluated
from ∑½m_{i}v_{i}^{2}.

Solution:

- Concepts:

The moment of inertia of a system about an axis - Reasoning:

The moment of inertia is I = ∑m_{i}r_{i}^{2}.

Here r_{i}is the perpendicular distance of particle i from the x-axis. - Details of the calculations:

(a) I = 4 kg 9 m^{2 }+ 2 kg 4 m^{2 }+ 3 kg 16 m^{2 }= 92 kgm^{2}.

The rotational kinetic energy is T_{rot}= ½Iω^{2 }= 46*4/s^{2 }= 184 J.

(b) The linear speed of particle i is v_{i }= ωr_{i}.

The linear speed of the 4 kg mass is v = 6 m/s, and its kinetic energy is ½mv^{2 }= 72 J.

The linear speed of the 2 kg mass is v = 4 m/s, and its kinetic energy is ½mv^{2 }= 16 J.

The linear speed of the 3 kg mass is v = 8 m/s, and its kinetic energy is ½mv^{2 }= 96 J.

The sum of the kinetic energies of the three particles is 184 J.

Two circular metal disks have the same mass M and the same thickness d. Disk
1 has a uniform density ρ_{1} which is less than ρ_{2}, the
uniform density of disk 2. Which disk, if either, has the larger moment
of inertia about its symmetry axis perpendicular to the plane of the disk?

Solution:

- Concepts:

The moment of inertia - Reasoning:

Let the radii of the disks be R_{1}and R_{2}respectively.

The disks have the same mass and thickness. Therefore ρ_{1}R_{1}^{2}= ρ_{2}R_{2}^{2}.

The moments of inertia are I_{1}= MR_{1}^{2}/2 and I_{2}= MR_{2}^{2}/2. - Details of the calculation:

I_{1}/I_{2}= R_{1}^{2}/R_{2}^{2}= ρ_{2}/ρ_{1}.

Since ρ_{1}< ρ_{2}, I_{1}> I_{2}. Disk 1 has the larger moment of inertia.

Consider a rigid body consisting of a collection of point masses m_{k}
at positions **r**_{k} in a body-fixed coordinate system. The system
rotates with angular velocity
**Ω**
about an axis which passes through the origin of the space-fixed coordinate
system. The angular momentum of the system about the origin is
**L** = Σ_{k}**r**_{k}×**p**_{k} = Σ_{k}m_{k}**r**_{k}×**v**_{k}
= Σ_{k}m_{k}**r**_{k}×(**Ω** × **r**_{k}).

(a) Define the Cartesian components I_{ij} (i, j = x, y, z) of
the inertia tensor, so that L_{i} = Σ_{j}I_{ij}Ω_{j},
where the Ω_{j} are the components of **Ω** along the
body-fixed axes.**
**(b) Now assume a continuous mass distribution of uniform mass density
ρ in a volume V having angular velocity

Define the Cartesian components I

Solution:

- Concepts:

The moment of inertia tensor. - Reasoning:

We are asked to define the moment of inertia tensor for a discrete and a continuous mass distribution. - Details of the calculation:

(a)**L**= Σ_{k}m_{k}**r**_{k}×(**Ω**×**r**_{k}) = Σ_{k}m_{k}[r_{k}^{2}**Ω**-**r**_{k}(**r**_{k}**·Ω**)].

[**A**×(**B**×**C**) = (**A**·**C**)**B**- (**A**·**B**)**C**]

We write this expression in component form.

L_{i}= Σ_{k}m_{k}[r_{k}^{2}Ω_{i}- (x_{k})_{i}(**r**_{k}**·Ω**)] = Σ_{k}m_{k}[Σ_{l}(x_{k})_{l}^{2}Ω_{i}- (x_{k})_{i}Σ_{j}(x_{k})_{j}]Ω_{j}], or

L_{i}= Σ_{j}Σ_{k}m_{k}[Σ_{l}(x_{k})_{l}^{2}δ_{ij}- (x_{k})_{i}(x_{k})_{j}]Ω_{j}= Σ_{j}I_{ij}Ω_{j }I_{ij}= Σ_{k}m_{k}[Σ_{l}(x_{k})_{l}^{2}δ_{ij}- (x_{k})_{i}(x_{k})_{j}].

I_{xx}= Σ_{k}m_{k}(y_{k}^{2}+ z_{k}^{2}), I_{yy}= Σ_{k}m_{k}(x_{k}^{2}+ z_{k}^{2}), I_{zz}= Σ_{k}m_{k}(x_{k}^{2}+ y_{k}^{2}),

I_{xy}= I_{yx}= -Σ_{k}m_{k}x_{k}y_{k}, I_{xz}= I_{zx}= -Σ_{k}m_{k}x_{k}z_{k}, I_{yz}= I_{zy}= -Σ_{k}m_{k}y_{k}z_{k}.

(b) For a continuous mass distribution of uniform mass density ρ**L**= ρ∫dV**[**r^{2}**Ω**-**r**(**r·Ω**)], and

I_{xx}= ρ∫dV (y^{2}+ z^{2}), I_{yy}= ρ∫dV(x^{2}+ z^{2}), I_{zz}= ρ∫dV(x^{2}+ y^{2}),

I_{xy}= I_{yx}= -ρ∫dVxy, I_{xz}= I_{zx}= -ρ∫dVxz, I_{yz}= I_{zy}= -ρ∫dVyz.

A cylinder with radius r, height h, and mass M has uniform mass distribution.

(a) Find the Cartesian components I_{ij} of its moment of inertia
tensor in the body-fixed coordinate system shown in the figure.

(b) If the cylinder is displaced along the z-axis so that its center of
mass is at the origin, find the components I_{ij} in this new coordinate
system.

Solution:

- Concepts:

The moment of inertia tensor. - Reasoning:

For a continuous mass distribution of uniform mass density ρ

I_{xx}= ρ∫dV (y^{2}+ z^{2}), I_{yy}= ρ∫dV(x^{2}+ z^{2}), I_{zz}= ρ∫dV(x^{2}+ y^{2}),

I_{xy}= I_{yx}= -ρ∫dVxy, I_{xz}= I_{zx}= -ρ∫dVxz, I_{yz}= I_{zy}= -ρ∫dVyz. - Details of the calculation:

(a) Here ρ = M/(πr^{2}h) = constant.

From symmetry I_{xy}= I_{yx}= I_{xz}= I_{zx}= I_{yz}= I_{zy}= 0. The coordinate axes are principal axes.

I_{3}= I_{zz}= ρ∫_{0}^{h}dz∫_{0}^{r}dr' 2πr'^{3}= 2πρhr^{4}/4 = Mr^{2}/2.

From symmetry I_{xx}= I_{yy}or I_{1}= I_{2}.

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

I_{1}= Mr^{2}/4 + [M/(πr^{2}h)]πr^{2}∫_{0}^{h}dz z^{2}= Mr^{2}/4 + Mh^{2}/3.

(b) The off-diagonal elements of the inertia tensor are still zero from symmetry.

Parallel axis theorem: I'_{ii}= I_{ii }+ M(a^{2}- a_{i}^{2}) = I_{ii }+ Ma_{⊥}^{2}where I_{ii}refers to a body-fixed coordinate system with the origin at the CM and I'_{ij }refers to a coordinate system with parallel axes and a different origin.

I_{zz}= I'_{zz}= Mr^{2}/2. I_{xx}= I_{yy}= I'_{xx}- Mh^{2}/4 = Mr^{2}/4 + Mh^{2}/12.

Find the fraction of the kinetic energy that is
translational and rotational when

(a) a hoop

(b) a disc and

(c) a sphere rolls down an inclined plane of height h. Find the velocity at
the bottom in each case. Compare with a block sliding without friction down the
plane.

A uniform rectangular object of mass m with sides a and b (b > a) and negligible thickness rotates with constant angular velocity ω about a diagonal through the center. Ignore gravity.

(a) What are the principal axes and principal moments of inertia?

(b) What is the angular momentum vector in the body coordinate system?

(c) What external torque must be applied to keep the object rotating with
constant angular velocity about the diagonal?