ENSC2004 - Lecture 19

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ENSC2004 - Lecture 18
Planar Kinetics - Slides

Equations of Translational Motion

• Translational motion:
  $\sum F_x=m(a_G{)}_x\text{and}\sum F_y=m(a_G{)}_y$
  The sum of external forces equals the mass times acceleration of the mass centre ($G$)

Equations of Rotational Motion

• Moment around point P can be written as:
  $\sum (r_i\times F_i)+\sum M_i=\bar{r}\times m{a}_G+I_G\alpha$
  $\sum (r_i\times F_i)$ $-$ moments from external forces
  $\sum M_i$ $-$ external couple moments
  $\text{or}\sum M_P=\sum ({\mathcal{M}}_k{)}_P$
  $\sum ({\mathcal{M}}_k{)}_P$ $-$ the kinetic moment about P
  $\bar{r}$ $-$ distance/position vector from $G$
  $m{a}_G$ $-$ inertial vector
  $I_G\alpha$ $-$ mass moment of inertia about $G$

• If point P is G, then we get scalar equation:
  $\sum M_G=I_G\alpha$
  The sum of moments about the centre of mass equals the moment of inertia about G times angular acceleration ($\alpha$)

Translation-Only Motion

• When a body undergoes only translation:
  • All particles share same acceleration → $a_G=a$
  • Angular acceleration $\alpha =0$
• Thus: $\sum F_x=m(a_G{)}_x,\sum F_y=m(a_G{)}_y,\sum M_G=0$
• Moments can be taken about other points if convenient:
  $\sum M_A=(m{\alpha }_G)d$

Curvilinear Translation

• Use n-t coordinate system
  $\sum F_n=m(a_G{)}_n,\sum F_t=m(a_G{)}_t,\sum M_G=0$
• Additionally:
  $\sum M_B=e[m(a_G{)}_t]-h[m(a_G{)}_n]$

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Fixed Axis Rotation - Slides

Fixed Axis Rotational Motion

• When a rigid body rotates about a fixed axis:
  • Every point moves in a circular path
  • The centre of mass $G$ follows a circle of radius $r_G$
  • Acceleration of $G$ has:
    • Tangential component: $(a_G{)}_t=r_G\alpha$
    • Normal component: $(a_G{)}_n=r_G{\omega }^2$
• We get equations of motion:
  $\sum F_n=m(a_G{)}_n=m{r}_G{\omega }^2,\sum F_t=m(a_G{)}_t=m{r}_G\alpha ,\sum M_O=I_O\alpha$
  $I_O$ $-$ moment of inertia about the fixed axis ($O$)
  Derived from parallel axis theorem
  $\alpha$ $-$ angular acceleration
  $\omega$ angular velocity (in radians)
  $m$ $-$ mass
  $I_O=I_G+m(r_G{)}^2$
  Thus:
  $\sum M_O=(I_G+m{r}_G^2)\alpha$

See: MMI

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ENSC2004 - Lecture 20