up

my dynamics cheat sheet

Nasser M. Abbasi, updated 10/2011

Contents

• 1 Miscellaneous hints
• 2 Formulas
• 3 Velocity and acceleration diagrams
  • 3.1 Spring pendulum
  • 3.2 pendulum with blob moving in slot
  • 3.3 spring pendulum with block moving in slot
  • 3.4 double pendulum
  
  
• 4 Velocity and acceleration of rigid body 2D
• 5 Velocity and acceleration of rigid body 3D
  • 5.1 Using Vehicle dynamics notations
  • 5.2 3D Not Using vehicle dynamics notations
    • 5.2.1 Derivation for $F=ma$ in 3D
    • 5.2.2 Derivation for $\tau=I\omega$ in 3D
    • 5.2.3 Derivation for $\tau=I\omega$ in 3D using principle axes
  
  
• 6 Wheel spining precession
• 7 unforced response of one degree of freedom linear system

1 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 Formulas

3 Velocity and acceleration diagrams

3.1 Spring pendulum

3.2 pendulum with blob moving in slot

3.3 spring pendulum with block moving in slot

3.4 double pendulum

4 Velocity and acceleration of rigid body 2D

5 Velocity and acceleration of rigid body 3D

5.1 Using Vehicle dynamics notations

5.2 3D Not Using vehicle dynamics notations

5.2.1 Derivation for $F=ma$ in 3D

$$\mathbf{F} =\frac{d}{dt}\mathbf{p}$$

$$=\frac{d}{dt}\left( m\mathbf{v}\right)$$

$$\mathbf{=}m\frac{d}{dt}\mathbf{v}$$

$$=m\left[ \begin{pmatrix} a_{x}\\ a_{y}\\ a_{z}% \end{pmatr... ...x} \otimes \begin{pmatrix} v_{x}\\ v_{y}\\ v_{z}% \end{pmatrix} \right]$$

$$=m\left[ \begin{pmatrix} a_{x}\\ a_{y}\\ a_{z}% \end{pmatr... ...x} & \omega_{y} & \omega_{z}\\ v_{x} & v_{y} & v_{z}% \end{vmatrix} \right]$$

$$=m\left[ \begin{pmatrix} a_{x}\\ a_{y}\\ a_{z}% \end{pmatr... ...a_{z}v_{x}\right) \\ \omega_{x}v_{y}-\omega_{y}v_{x}% \end{pmatrix} \right]$$

$$=m\begin{pmatrix} a_{x}+\omega_{y}v_{z}-\omega_{z}v_{y}\\ a_{y... ...v_{z}+\omega_{z}v_{x}\\ a_{z}+\omega_{x}v_{y}-\omega_{y}v_{x}% \end{pmatrix}$$

5.2.2 Derivation for $\tau=I\omega$ in 3D

Let $A=I\omega$ then using the rule

$$\mathbf{\tau} =\left( \frac{d}{dt}\mathbf{A}\right)$$

$$=\left( \frac{d}{dt}\mathbf{A}\right) _{\text{resolved}}+\mathbf{\omega \times A}%$$

Then $\tau=I\omega$ can be found for the general case

$$\mathbf{\tau} =\frac{d}{dt}\left[ \overset{\mathbf{A}}{\overbrace{% \begin{pma... ... \begin{pmatrix} \omega_{x}\\ \omega_{y}\\ \omega_{z}% \end{pmatrix} }}$$

$$=% \begin{pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy}... ...mega_{z}\\ I_{zx}\omega_{x}+I_{yz}\omega_{y}+I_{zz}\omega_{z}% \end{pmatrix}$$

$$=% \begin{pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy}... ...left( I_{zx}\omega_{x}+I_{yz}\omega_{y}+I_{zz}\omega_{z}\right) \end{vmatrix}$$

$$=% \begin{pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy}... ...\left( I_{xx}\omega_{x}+I_{xy}\omega_{y}+I_{xz}\omega_{z}\right) \end{pmatrix}$$

5.2.3 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

$$\mathbf{\tau} =\frac{d}{dt}\left[ \overset{\mathbf{A}}{\overbrace{% \begin{pma... ... \begin{pmatrix} \omega_{x}\\ \omega_{y}\\ \omega_{z}% \end{pmatrix} }}$$

$$=% \begin{pmatrix} I_{xx} & 0 & 0\\ 0 & I_{yy} & 0\\ 0 & 0 ... ...ix} I_{xx}\omega_{x}\\ I_{yy}\omega_{y}\\ I_{zz}\omega_{z}% \end{pmatrix}$$

$$=% \begin{pmatrix} I_{xx}\alpha_{x}\\ I_{yy}\alpha_{y}\\ I_... ..._{z}\\ I_{xx}\omega_{x} & I_{yy}\omega_{y} & I_{zz}\omega_{z}% \end{vmatrix}$$

$$=% \begin{pmatrix} I_{xx}\alpha_{x}\\ I_{yy}\alpha_{y}\\ I_... ...yy}\omega_{y}\right) -\omega_{y}\left( I_{xx}\omega _{x}\right) \end{pmatrix}$$

$$=% \begin{pmatrix} I_{xx}\alpha_{x}\\ I_{yy}\alpha_{y}\\ I_... ...{zz}\right) \\ \omega_{x}\omega_{y}\left( I_{yy}-I_{xx}\right) \end{pmatrix}$$

So, we can see how much simpler it became when using principle axes. Compare the above to

$$% \begin{pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{yx} & I_{yy} ... ...eft( I_{xx}\omega_{x}+I_{xy}\omega_{y}+I_{xz}\omega_{z}\right) \end{pmatrix}$$

So, always use principle axes for the body fixed coordinates system!

6 Wheel spining precession

7 unforced response of one degree of freedom linear system