## Chapter 2Dynamics equations, kinematics, velocity and acceleration diagrams

### 2.1 Derivation of rotation formula

This formula is very important. Will show its derivation now in details. It is how to express vectors in rotating frames.

Consider this diagram

In the above, the small axis $$x,y$$ is a frame attached to some body which rotate around this axis with angular velocity $$\omega$$ (measured by the inertial frame of course). All laws derived below are based on the following one rule\begin {equation} \left . \frac {d}{dt}r\right \vert _{absolute}=\left . \frac {d}{dt}r\right \vert _{relative}+\omega \times r \tag {1} \end {equation}

Lets us see how to apply this rule. Let us express the position vector of the particle $$r_{p}$$. We can see by normal vector additions that the position vector of particle is\begin {equation} r_{p}=r_{o}+r \tag {2} \end {equation}

Notice that nothing special is needed here, since we have not yet looked at rate of change with time. The complexity (i.e. using rule (1)) appears only when we want to look at velocities and accelerations. This is when we need to use the above rule (1). Let us now ﬁnd the velocity of the particle. From above$\dot {r}_{p}=\dot {r}_{o}+\dot {r}$

Every time we take derivatives, we stop and look. For any vector that originates from the moving frame, we must apply rule (1) to it. That is all. In the above, only $$r$$ needs rule (1) applied to it, since that is the only vector measure from the moving frame. Replacing $$\dot {r}_{p}$$ by $$V_{p}$$ and $$\dot {r}_{o}$$ by $$V_{o}$$, meaning the velocity of $$P$$ and $$o$$, Hence the above becomes$V_{p}=V_{o}+\dot {r}$

and now we apply rule (1) to expand $$\dot {r}$$\begin {equation} V_{p}=V_{o}+\left ( V_{rel}+\omega \times r\right ) \tag {3} \end {equation}

where $$V_{rel}$$ is just $$\left . \frac {d}{dt}r\right \vert _{relative}\,$$

The above is the ﬁnal expression for the velocity of the particle $$V_{p}$$ using its velocity as measured by the moving frame in order to complete the expression.

So the above says that the absolute velocity of the particle is equal to the absolute velocity of the base of the moving frame + something else and this something else was $$\left ( V_{rel}+\omega \times r\right )$$

Now we will ﬁnd the absolute acceleration of $$P$$. Taking time derivatives of (3) gives\begin {equation} \dot {V}_{p}=\dot {V}_{o}+\left ( \dot {V}_{rel}+\dot {\omega }\times r+\omega \times \dot {r}\right ) \tag {4} \end {equation}

As we said above, each time we take time derivatives, we stop and look for vectors which are based on the moving frame, and apply rule (1) to them. In the above, $$\dot {V}_{rel}$$ and $$\dot {r}$$ qualify. Apply rule (1) to $$\dot {V}_{rel}$$ gives\begin {equation} \dot {V}_{rel}=a_{rel}+\omega \times V_{rel} \tag {5} \end {equation}

where $$a_{rel}$$ just means the acceleration relative to moving frame. And applying rule (1) to $$\dot {r}$$ gives\begin {equation} \dot {r}=V_{rel}+\omega \times r \tag {6} \end {equation}

Replacing (5) and (6) into (4) gives\begin {align} a_{p} & =a_{o}+\left ( a_{rel}+\omega \times V_{rel}+\dot {\omega }\times r+\omega \times \left ( V_{rel}+\omega \times r\right ) \right ) \nonumber \\ & =a_{o}+a_{rel}+\left ( \omega \times V_{rel}\right ) +\left ( \dot {\omega }\times r\right ) +\left ( \omega \times V_{rel}\right ) +\left ( \omega \times \left ( \omega \times r\right ) \right ) \nonumber \\ & =a_{o}+a_{rel}+2\left ( \omega \times V_{rel}\right ) +\left ( \dot {\omega }\times r\right ) +\left ( \omega \times \left ( \omega \times r\right ) \right ) \tag {7} \end {align}

Eq (7) says that the absolute acceleration $$a_{p}$$ of $$P$$ is the sum of the acceleration of the base $$a_{o}$$ of the moving frame plus the relative acceleration $$a_{rel}$$ of the particle to the moving frame plus $$2\left ( \omega \times V_{rel}\right ) +\left ( \dot {\omega }\times r\right ) +\left ( \omega \times \left ( \omega \times r\right ) \right )$$

Hence, using Eq(3) and Eq(7) gives us the expressions we wanted for velocity and acceleration.

### 2.2 Miscellaneous hints

1. When ﬁnding the generalized force for the user with the Lagrangian method (the hardest step), using the virtual work method, if the force (or virtual work by the force) ADDS energy to the system, then make the sign of the force positive otherwise the sign is negative.
2. For damping force, the sign is always negative.
3. External forces such as linear forces applied, torque applied, in general, are positive.
4. Friction force is negative (in general) as friction takes energy from the system like damping.

### 2.5 Velocity and acceleration of rigid body 2D

Finding linear acceleration of center of mass of a rigid body under pure rotation using ﬁxed body coordinates.

In the above $$U$$ is the speed of the center of mass in the direction of the $$x$$ axis, where this axis is ﬁxed on the body itself. Similarly, $$V$$ is the speed of the center of mass in the direction of the $$y$$ axis, where the $$y$$ axis is attached to the body itself.

Just remember that all these speeds (i.e. $$U$$,$$V$$) and accelerations ($$a_{x}$$, $$a_{y}$$) are still being measured by an observer in the inertial frame. It is only that the directions of the velocity components of the center of mass is along an axis ﬁxed on the body. Only the direction. But actual speed measurements are still done by a stationary observer. Since clearly if the observer was sitting on the body itself, then they will measure the speeds to be zero in that case.

### 2.6 Velocity and acceleration of rigid body 3D

#### 2.6.2 3D Not Using vehicle dynamics notations

##### 2.6.2.1 Derivation for $$F=ma$$ in 3D

\begin {align*} \boldsymbol {F} & =\frac {d}{dt}\boldsymbol {p}\\ & =\frac {d}{dt}\left ( m\boldsymbol {v}\right ) \\ & \boldsymbol {=}m\frac {d}{dt}\boldsymbol {v}\\ & =m\left [ \begin {pmatrix} a_{x}\\ a_{y}\\ a_{z}\end {pmatrix} +\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} \otimes \begin {pmatrix} v_{x}\\ v_{y}\\ v_{z}\end {pmatrix} \right ] \\ & =m\left [ \begin {pmatrix} a_{x}\\ a_{y}\\ a_{z}\end {pmatrix} +\det \begin {vmatrix} \boldsymbol {i} & \boldsymbol {j} & \boldsymbol {k}\\ \omega _{x} & \omega _{y} & \omega _{z}\\ v_{x} & v_{y} & v_{z}\end {vmatrix} \right ] \\ & =m\left [ \begin {pmatrix} a_{x}\\ a_{y}\\ a_{z}\end {pmatrix} +\begin {pmatrix} \omega _{y}v_{z}-\omega _{z}v_{y}\\ -\left ( \omega _{x}v_{z}-\omega _{z}v_{x}\right ) \\ \omega _{x}v_{y}-\omega _{y}v_{x}\end {pmatrix} \right ] \\ & =m\begin {pmatrix} a_{x}+\omega _{y}v_{z}-\omega _{z}v_{y}\\ a_{y}-\omega _{x}v_{z}+\omega _{z}v_{x}\\ a_{z}+\omega _{x}v_{y}-\omega _{y}v_{x}\end {pmatrix} \end {align*}

##### 2.6.2.2 Derivation for $$\tau =I\omega$$ in 3D

Let $$A=I\omega$$ then using the rule\begin {align*} \boldsymbol {\tau } & =\left ( \frac {d}{dt}\boldsymbol {A}\right ) \\ & =\left ( \frac {d}{dt}\boldsymbol {A}\right ) _{\text {resolved}}+\boldsymbol {\omega \times A} \end {align*}

Then $$\tau =I\omega$$  can be found for the general case\begin {align*} \boldsymbol {\tau } & =\frac {d}{dt}\left [ \overset {\boldsymbol {A}}{\overbrace {\begin {pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} & I_{yz}\\ I_{zx} & I_{yz} & I_{zz}\end {pmatrix}\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} }}\right ] +\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} \times \overset {\boldsymbol {A}}{\overbrace {\begin {pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} & I_{yz}\\ I_{zx} & I_{yz} & I_{zz}\end {pmatrix}\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} }}\\ & =\begin {pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} & I_{yz}\\ I_{zx} & I_{yz} & I_{zz}\end {pmatrix}\begin {pmatrix} \alpha _{x}\\ \alpha _{y}\\ \alpha _{z}\end {pmatrix} +\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} \times \begin {pmatrix} I_{xx}\omega _{x}+I_{xy}\omega _{y}+I_{xz}\omega _{z}\\ I_{yx}\omega _{x}+I_{yy}\omega _{y}+I_{yz}\omega _{z}\\ I_{zx}\omega _{x}+I_{yz}\omega _{y}+I_{zz}\omega _{z}\end {pmatrix} \\ & =\begin {pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} & I_{yz}\\ I_{zx} & I_{yz} & I_{zz}\end {pmatrix}\begin {pmatrix} \alpha _{x}\\ \alpha _{y}\\ \alpha _{z}\end {pmatrix} +\det \begin {vmatrix} \boldsymbol {i} & \boldsymbol {j} & \boldsymbol {k}\\ \omega _{x} & \omega _{y} & \omega _{z}\\ \left ( I_{xx}\omega _{x}+I_{xy}\omega _{y}+I_{xz}\omega _{z}\right ) & \left ( I_{yx}\omega _{x}+I_{yy}\omega _{y}+I_{yz}\omega _{z}\right ) & \left ( I_{zx}\omega _{x}+I_{yz}\omega _{y}+I_{zz}\omega _{z}\right ) \end {vmatrix} \\ & =\begin {pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} & I_{yz}\\ I_{zx} & I_{yz} & I_{zz}\end {pmatrix}\begin {pmatrix} \alpha _{x}\\ \alpha _{y}\\ \alpha _{z}\end {pmatrix} +\begin {pmatrix} \omega _{y}\left ( I_{zx}\omega _{x}+I_{yz}\omega _{y}+I_{zz}\omega _{z}\right ) -\omega _{z}\left ( I_{yx}\omega _{x}+I_{yy}\omega _{y}+I_{yz}\omega _{z}\right ) \\ \omega _{x}\left ( I_{zx}\omega _{x}+I_{yz}\omega _{y}+I_{zz}\omega _{z}\right ) -\omega _{z}\left ( I_{xx}\omega _{x}+I_{xy}\omega _{y}+I_{xz}\omega _{z}\right ) \\ \omega _{x}\left ( I_{yx}\omega _{x}+I_{yy}\omega _{y}+I_{yz}\omega _{z}\right ) -\omega _{y}\left ( I_{xx}\omega _{x}+I_{xy}\omega _{y}+I_{xz}\omega _{z}\right ) \end {pmatrix} \end {align*}

##### 2.6.2.3 Derivation for $$\tau =I\omega$$ in 3D using principle axes

The above derivation simpliﬁes now since we will be using principle axes. In this case, all cross products of moments of inertia vanish.$I=\begin {pmatrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0 & 0 & I_{zz}\end {pmatrix}$

Hence\begin {align*} \boldsymbol {\tau } & =\frac {d}{dt}\left [ \overset {\boldsymbol {A}}{\overbrace {\begin {pmatrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0 & 0 & I_{zz}\end {pmatrix}\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} }}\right ] +\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} \times \overset {\boldsymbol {A}}{\overbrace {\begin {pmatrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0 & 0 & I_{zz}\end {pmatrix}\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} }}\\ & =\begin {pmatrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0 & 0 & I_{zz}\end {pmatrix}\begin {pmatrix} \alpha _{x}\\ \alpha _{y}\\ \alpha _{z}\end {pmatrix} +\begin {pmatrix} \omega _{x}\\ \omega _{y}\\ \omega _{z}\end {pmatrix} \times \begin {pmatrix} I_{xx}\omega _{x}\\ I_{yy}\omega _{y}\\ I_{zz}\omega _{z}\end {pmatrix} \\ & =\begin {pmatrix} I_{xx}\alpha _{x}\\ I_{yy}\alpha _{y}\\ I_{zz}\alpha _{z}\end {pmatrix} +\det \begin {vmatrix} \boldsymbol {i} & \boldsymbol {j} & \boldsymbol {k}\\ \omega _{x} & \omega _{y} & \omega _{z}\\ I_{xx}\omega _{x} & I_{yy}\omega _{y} & I_{zz}\omega _{z}\end {vmatrix} \\ & =\begin {pmatrix} I_{xx}\alpha _{x}\\ I_{yy}\alpha _{y}\\ I_{zz}\alpha _{z}\end {pmatrix} +\begin {pmatrix} \omega _{y}\left ( I_{zz}\omega _{z}\right ) -\omega _{z}\left ( I_{yy}\omega _{y}\right ) \\ -\omega _{x}\left ( I_{zz}\omega _{z}\right ) +\omega _{z}\left ( I_{xx}\omega _{x}\right ) \\ \omega _{x}\left ( I_{yy}\omega _{y}\right ) -\omega _{y}\left ( I_{xx}\omega _{x}\right ) \end {pmatrix} \\ & =\begin {pmatrix} I_{xx}\alpha _{x}\\ I_{yy}\alpha _{y}\\ I_{zz}\alpha _{z}\end {pmatrix} +\begin {pmatrix} \omega _{y}\omega _{z}\left ( I_{zz}-I_{yy}\right ) \\ \omega _{x}\omega _{z}\left ( I_{xx}-I_{zz}\right ) \\ \omega _{x}\omega _{y}\left ( I_{yy}-I_{xx}\right ) \end {pmatrix} \end {align*}

So, we can see how much simpler it became when using principle axes. Compare the above to$\begin {pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} & I_{yz}\\ I_{zx} & I_{yz} & I_{zz}\end {pmatrix}\begin {pmatrix} \alpha _{x}\\ \alpha _{y}\\ \alpha _{z}\end {pmatrix} +\begin {pmatrix} \omega _{y}\left ( I_{zx}\omega _{x}+I_{yz}\omega _{y}+I_{zz}\omega _{z}\right ) -\omega _{z}\left ( I_{yx}\omega _{x}+I_{yy}\omega _{y}+I_{yz}\omega _{z}\right ) \\ \omega _{x}\left ( I_{zx}\omega _{x}+I_{yz}\omega _{y}+I_{zz}\omega _{z}\right ) -\omega _{z}\left ( I_{xx}\omega _{x}+I_{xy}\omega _{y}+I_{xz}\omega _{z}\right ) \\ \omega _{x}\left ( I_{yx}\omega _{x}+I_{yy}\omega _{y}+I_{yz}\omega _{z}\right ) -\omega _{y}\left ( I_{xx}\omega _{x}+I_{xy}\omega _{y}+I_{xz}\omega _{z}\right ) \end {pmatrix}$ So, always use principle axes for the body ﬁxed coordinates system!

### 2.8 References

1. Structural Dynamics 5th edition. Mario Paz, William Leigh

### 2.9 Misc. items

The Jacobian matrix for a system of diﬀerential equations, such as\begin {align*} x^{\prime }\left ( t\right ) & =f\left ( x,y,z\right ) \\ y^{\prime }\left ( t\right ) & =g\left ( x,y,z\right ) \\ z^{\prime }\left ( t\right ) & =h\left ( x,y,z\right ) \end {align*}

is given by$J=\begin {pmatrix} \frac {df}{dx} & \frac {df}{dy} & \frac {df}{dz}\\ \frac {dg}{dx} & \frac {dg}{dy} & \frac {dg}{dz}\\ \frac {dh}{dx} & \frac {dh}{dy} & \frac {dh}{dz}\end {pmatrix}$

For example, for the given the following 3 set of coupled diﬀerential equations in $$n^{3}$$\begin {align*} x^{\prime }\left ( t\right ) & =-y\left ( t\right ) -z\left ( t\right ) \\ y^{\prime }\left ( t\right ) & =x\left ( t\right ) +ay\left ( t\right ) \\ z^{\prime }\left ( t\right ) & =b+z\left ( t\right ) \left ( x\left ( t\right ) -c\right ) \end {align*}

then the Jacobian matrix is$J=\begin {pmatrix} 0 & -1 & -1\\ 1 & a & 0\\ z\left ( t\right ) & 0 & x\left ( t\right ) -c \end {pmatrix}$

Now to ﬁnd stability of this system, we evaluate this matrix at $$t=t_{0}$$ where $$x\left ( t_{0}\right ) ,y\left ( t_{0}\right ) ,z\left ( t_{0}\right )$$ is a point in this space (may be stable point or initial conditions, etc...) and then $$J$$ become all numerical now. Then we can evaluate the eigenvalues of the resulting matrix and look to see if all eigenvalues are negative. If so, this tells us that the point is a stable point. I.e. the system is stable.

If X is $$N(0,1)$$ distributed then $$mu+sigma\ast X$$ is $$N(mu,sigma^{2})$$ distributed.