MECH2004 - Lecture 2

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MECH2004 - Lecture 1
Kinetics of Particles - Slides

Curvilinear Motion/Force

Newton’s Laws of Motion

• In dynamics, we commonly use Newton’s 2nd law:
  $F=ma$

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• Rectangular coordinates:
  $a_x=\ddot{x}{a}_y=\ddot{y}$
• Normal and tangential coordinates:
  $a_n=\rho {\dot{\beta }}^2=v^2/\rho =v\dot{\beta }$
  $a_t=\dot{v}v=\rho \dot{\beta }$
• Polar coordinates:
  $a_r=\ddot{r}-r{\dot{\theta }}^2$
  $a_{\theta }=r\ddot{\theta }+2\dot{r}\dot{\theta }$

Work

Work of a Force

• Work is the product of force and the displacement components in the direction of the force $dU=\text{F}ds\cos \theta$
  • Where $\theta$ is the angle between the force and displacement vector
    • $dU$ is the increment of work done

Integrate and use dot product to get:
$U_{1-2}={\int }_{r_1}^{r_2}\text{F}\cdot dr$
If $\text{F}$ is a function of position, this becomes:
$U_{1-2}={\int }_{r_1}^{r_2}\text{F}\cos \theta ds$
If both $\text{F}$ and $\theta$ are constant, the equation can be simplified to:
$U_{1-2}={\text{F}}_c\cos \theta (s_2-s_1)$

Work is positive if the force and the movement are in the same direction If the force and displacement are perpendicular, the work is zero

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Work of a Spring Force

• When stretched, a linear elastic spring develops force ${\text{F}}_s=ks$
  $U_s=-0.5k[(s_2{)}^2-(s_1{)}^2]$
  ${\text{F}}_s$ is exerted on a particle in the opposite direction to the force exerted on the spring, meaning the work done on the particle will be negative

1. Equation above only for linear springs
2. Work of a spring is not just spring force times distance at some point
3. Always double check the sign of the spring work!

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Work of a Weight

• The work done by gravity acting on a particle is given by:
  $U_w={\int }_{y_1}^{y_2}-\text{W}dy=-\text{W}\Delta y$
  If $\Delta y$ is upward, the work is negative since weight force acts downwards

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Energy

Kinetic Energy

• When we considered kinetic energy of a particle, there was no rotation
  $T=\frac{1}{2}m{v}^2$
• However, for a rigid body, we need to account for rotational velocity
  $T=\frac{1}{2}m(v_G{)}^2+\frac{1}{2}{I}_G{\omega }^2$
  Remember that $G$ means a vector is applied at the centre of mass

We can simplify this equation depending on the scenario:

1. Pure Translation $-$ body subject to only curvilinear or rectilinear translation (no rotational kinetic energy)
   $T=\frac{1}{2}m(v_G{)}^2$

2. Pure Rotation $-$ body rotates about fixed axis passing through $O$
   $T=\frac{1}{2}[I_G+m(r_G{)}^2]{\omega }^2=\frac{1}{2}{I}_O{\omega }^2$
   Note: $v_G=r_G\omega$

   • If rotation occurs around $G$ (i.e. $O=G$), $v_G=0$
     $\therefore T=\frac{1}{2}{I}_G{\omega }^2$

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Principle of Work and Energy

• Recall: Sum of the initial kinetic energy and work done by all external forces and couple moments equals the body’s final kinetic energy
  $T_1+\sum U_{1-2}=T_2$

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Gravitational Potential Energy

• Potential energy is function of bodies’ height above/below a datum
  $V_g=W{y}_G$

  

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Elastic Potential Energy

• Potential energy of a spring:
  ${\text{V}}_e=\frac{1}{2}k{s}^2$
  

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Conservation of Energy

• When a particle is acted upon by only conservative forces, the sum of kinetic energy and potential energy remains constant
  $T_1+V_1=T_2+V_2=\text{Constant}$

Note: $T_n=0.5\text{m}(v_n{)}^2$

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Momentum

Principle of Impulse and Momentum

• Force → creates impulse → creates change in momentum
• Use these equations to solve problems with force, velocity, and time

Linear impulse-linear momentum equation:
${\mathbf{L}}_1+\sum {\int }_{t_1}^{t_2}\mathbf{F}dt={\mathbf{L}}_2$
$\mathbf{L}=\text{m}{\mathbf{v}}_{\text{G}}$

Angular impulse-angular momentum equation:
$({\mathbf{H}}_{\text{G}}{)}_1+\sum {\int }_{t_1}^{t_2}{\mathbf{M}}_{\text{G}}dt=({\mathbf{H}}_{\text{G}}{)}_2$
${\mathbf{H}}_{\text{G}}={\text{I}}_{\text{G}}\omega$

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Angular Momentum

• Angular momentum of a particle about $O$ is defined as the “moment” of the particles linear motion about $O$
  ${\text{H}}_O=r\times \text{m}v$

  
  $r$ in this diagram would be $d$

• Angular moment is not the same as the linear momentum!
  $\sum {\text{M}}_O=r\times F={\dot{\text{H}}}_O$
  A dot above a character represents a time derivative

• If something starts at rest, its initial angular momentum = 0

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Principle of Angular Impulse and Momentum

By integrating the equation above, we can get:
$(H_O{)}_1+\sum {\int }_{t_1}^{t_2}{M}_{O}dt=(H_O{)}_2$

• This is known as the principle of angular impulse and momentum
  • $\sum {\int }_{t_1}^{t_2}{M}_{O}dt$ → angular impulse
• Angular momentum is sometimes represented with $L$

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Conservation of Angular Momentum

• If sum of angular impulses is zero, angular momentum is conserved
  $(H_O{)}_1=(H_O{)}_2$
  e.g. if a particle is only subject to forces pointing towards $O$

  
  A good example of this in practice, would be the tension force acting upon a ball swinging around a pole from a rope

• Typically you will use: $0=m_a\cdot v_a\cdot r_a-m_b\cdot v_b\cdot r_b$

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MECH2004 - Lecture 3