Kinematics of Rigid Bodies

Types of plane motion

General plane motion can always be treated as a translation of a chosen base point plus a rotation about that point.

Relative velocity relation

Let A and B be two points of a rigid body having angular velocity $\omega$ (counter-clockwise positive, along $\mathbf{k}$).

        y
        |      B
        |     /
        |    /  r_B/A
        |  A
        | / r_A
        |/
        O------------ x

From the geometry: $\mathbf{r}_B = \mathbf{r}_A + \mathbf{r}_{B/A}$. Differentiating with respect to time:

$$\mathbf{v}_B = \mathbf{v}_A + \dot{\mathbf{r}}_{B/A}$$

In a rigid body the distance AB is constant, so the vector $\mathbf{r}_{B/A}$ can only rotate with angular velocity $\omega$ and $\dot{\mathbf{r}}_{B/A} = \boldsymbol{\omega}\times\mathbf{r}_{B/A}$. Hence

$$\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega}\times\mathbf{r}_{B/A}$$

The relative velocity $\mathbf{v}_{B/A}$ has magnitude $\omega\,(AB)$ and is perpendicular to AB, in the sense of $\omega$. Solving problems: draw the velocity triangle $\mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A}$ using known directions, or write components and solve.

Given: $OA = r = 0.2$ m, $AB = l = 0.8$ m, $N = 300$ rpm counter-clockwise, $\theta = 60^\circ$.

        A (pin)
       /\
      /  \  AB = 0.8
     /60  \
   O-------B-------> slider (horizontal guide)

Crank speed

$$\omega_{OA} = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s} \qquad v_A = \omega_{OA}\,r = 6.283\ \text{m/s}$$

$v_A$ is perpendicular to OA. Its components: $v_{Ax} = -v_A\sin60^\circ = -5.441$, $v_{Ay} = v_A\cos60^\circ = 3.142$ m/s.

Geometry of the rod

$\sin\phi = \dfrac{r\sin60^\circ}{l} = \dfrac{0.1732}{0.8} = 0.2165$, so the rod angle $\phi = 12.50^\circ$. Horizontal projection of AB: $d = l\cos\phi = 0.7810$ m.

Relative velocity method

$\mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A}$, with $\mathbf{v}_B$ horizontal and $\mathbf{v}_{B/A}$ perpendicular to AB (magnitude $\omega_{AB}\times l$).

Vertical components (B has none): $v_{Ay} = \omega_{AB}\,d$

$$\omega_{AB} = \frac{3.142}{0.7810} = 4.022\ \text{rad/s (clockwise)}$$

Horizontal components:

$$v_B = v_{Ax} - \omega_{AB}(AB\text{ vertical drop}) = -5.441 - (4.022)(0.1732) = -6.138\ \text{m/s}$$

The negative sign means the slider moves toward O (to the left).

Verification by instantaneous centre

I is the intersection of the perpendicular to $v_A$ at A (line OA extended) and the perpendicular to $v_B$ at B (vertical through B).

• $x_B = 0.2\cos60^\circ + 0.7810 = 0.8810$ m
• I lies on line OA at that x: $y_I = x_B\tan60^\circ = 1.5260$ m, so $IB = 1.526$ m
• $IA = \sqrt{(x_B - 0.1)^2 + (y_I - 0.1732)^2} = 1.562$ m

$$\omega_{AB} = \frac{v_A}{IA} = \frac{6.283}{1.562} = 4.022\ \text{rad/s}$$ $$v_B = \omega_{AB}\times IB = 4.022 \times 1.526 = 6.138\ \text{m/s}$$

The two methods agree.

Answer: Slider velocity = 6.14 m/s toward the crank centre; ω of connecting rod = 4.02 rad/s clockwise.

Given: $L = 4$ m, $\theta = 60^\circ$ with the floor, $v_A = 2$ m/s downward along the wall.

Locating the instantaneous centre

$v_A$ is vertical, so the perpendicular to it at A is the horizontal line through A. $v_B$ is horizontal, so the perpendicular at B is the vertical line through B. These meet at I.

   wall
    |A
    | \         I = instantaneous centre
  --+--\-----I     (horizontal from A,
    |   \G   |      vertical from B)
    |    \   |
    +-----\--B--- floor

Coordinates (origin at the wall-floor corner): A $(0,\ 3.464)$, B $(2,\ 0)$, so I $= (2,\ 3.464)$ m.

• $IA = L\cos60^\circ = 2.00$ m
• $IB = L\sin60^\circ = 3.464$ m
• $IG = L/2 = 2$ m (I, A, B and the corner form a rectangle, and G is the midpoint of the diagonal AB)

Angular velocity

$$\omega = \frac{v_A}{IA} = \frac{2}{2.00} = 1.00\ \text{rad/s (counter-clockwise)}$$

Velocities of B and G

$$v_B = \omega \times IB = 1.00 \times 3.464 = 3.464\ \text{m/s (horizontal, away from the wall)}$$ $$v_G = \omega \times IG = 1.00 \times 2 = 2.00\ \text{m/s}$$

$v_G$ is perpendicular to IG.

Answer: I is 2 m horizontally from A and 3.464 m vertically from B; ω = 1.00 rad/s; v_B = 3.46 m/s; v_G = 2.00 m/s.

Instantaneous centre (IC) of a body in plane motion is the point (on the body or its extension) whose velocity is zero at that instant. At that instant the body appears to rotate about it with angular velocity $\omega$, so the velocity of any point P is $v_P = \omega\,(IP)$, perpendicular to $IP$.

Methods of locating the IC

1. Directions of velocities of two points known (not parallel). Draw perpendiculars to $v_A$ at A and to $v_B$ at B. Their intersection is the IC (example: ends of a sliding ladder, or crank pin and slider of an engine).

      A ->v_A     perpendicular lines meet at I
      |
      |       B
      |      /|
      +-----I-+ 

2. Velocities parallel at A and B and both perpendicular to AB. Join the tips of the velocity vectors with a straight line; its intersection with the line AB gives the IC. If the vectors point the same way the IC lies outside AB (on the side of the smaller velocity); if they are opposite it lies between A and B. Because $v_A/IA = v_B/IB$, this follows from similar triangles.

3. Velocities parallel but not perpendicular to AB, equal in magnitude. The IC is at infinity: the body is in instantaneous translation ($\omega = 0$).

4. Rolling without slipping. The point of contact C with the fixed surface has zero velocity, so the IC is at the contact point. For a wheel of radius $r$: $\omega = v_O/r$, and the top point moves at $2v_O$.

Is the acceleration of the IC zero?

No. The IC has zero velocity but in general it has acceleration, so it must not be used to find accelerations as if it were a fixed point. For a rolling wheel, for example, the contact point has an acceleration $\omega^2 r$ directed toward the wheel centre.

Given: $r = 0.5$ m, $v_O = 6$ m/s, $a_O = 2$ m/s² (both forward, +x); rolling without slipping.

Angular motion

The wheel rolls forward, so the rotation is clockwise.

$$\omega = \frac{v_O}{r} = \frac{6}{0.5} = 12.0\ \text{rad/s} \qquad \alpha = \frac{a_O}{r} = \frac{2}{0.5} = 4.0\ \text{rad/s}^2$$

Relative acceleration equation

For any point P: $\mathbf{a}_P = \mathbf{a}_O + \mathbf{a}_{P/O}$, where $\mathbf{a}_{P/O}$ has a tangential part $\alpha r$ (perpendicular to OP, in the sense of $\alpha$) and a normal part $\omega^2 r$ directed from P toward O.

$$\omega^2 r = 12^2 \times 0.5 = 72.0\ \text{m/s}^2 \qquad \alpha r = 4 \times 0.5 = 2.0\ \text{m/s}^2$$

For a clockwise rotation, the tangential part at the bottom point points backward, at the top forward, at the front point downward.

(a) Contact point C (directly below O)

• $a_O = 2$ m/s² forward
• Tangential: $\alpha r = 2$ m/s² backward
• Normal: $\omega^2 r = 72$ m/s² upward (toward O)

$$a_{Cx} = 2 - 2 = 0 \qquad a_{Cy} = 72.0\ \text{m/s}^2 \Rightarrow a_C = 72.0\ \text{m/s}^2\ \text{upward}$$

(b) Top point T

• Tangential: 2 m/s² forward; normal: 72 m/s² downward

$$a_{Tx} = 2 + 2 = 4.0,\quad a_{Ty} = -72.0 \Rightarrow a_T = 72.11\ \text{m/s}^2$$

directed $\tan^{-1}(72/4) = 86.8^\circ$ below the horizontal.

(c) Front point F (level with O)

• Tangential: 2 m/s² downward; normal: 72 m/s² backward (toward O)

$$a_{Fx} = 2 - 72 = -70.0,\quad a_{Fy} = -2.0 \Rightarrow a_F = 70.03\ \text{m/s}^2$$

Answer: ω = 12 rad/s clockwise, α = 4 rad/s²; a_C = 72.0 m/s² (vertically up); a_T = 72.1 m/s²; a_F = 70.0 m/s².

Given: $r = OA = 0.15$ m, $l = AB = 0.6$ m, $\omega_{OA} = 20$ rad/s (constant, counter-clockwise), $\theta = 45^\circ$.

Geometry

$\sin\phi = \dfrac{0.15\sin45^\circ}{0.6} = 0.1768$, $\phi = 10.18^\circ$. Horizontal projection of AB: $d = l\cos\phi = 0.5906$ m. Vertical drop from A to B: $0.15\sin45^\circ = 0.1061$ m.

Velocity analysis

$v_A = \omega r = 20 \times 0.15 = 3$ m/s, perpendicular to OA: $\mathbf{v}_A = (-2.121,\ 2.121)$ m/s. B moves horizontally, so the vertical components give

$$\omega_{AB} = \frac{2.121}{0.5906} = 3.592\ \text{rad/s (clockwise)}$$

Acceleration analysis

Since $\omega_{OA}$ is constant, A has only a normal acceleration toward O:

$$a_A = \omega^2 r = 20^2 \times 0.15 = 60\ \text{m/s}^2,\quad \mathbf{a}_A = (-42.43,\ -42.43)\ \text{m/s}^2$$

Relative acceleration of B with respect to A has:

• normal part $\omega_{AB}^2\,l = (3.592)^2(0.6) = 7.74$ m/s² directed from B toward A,
• tangential part $\alpha_{AB}\,l$ perpendicular to AB.

Resolve $\mathbf{a}_B = \mathbf{a}_A + \mathbf{a}^n_{B/A} + \mathbf{a}^t_{B/A}$ with B having no vertical acceleration. The vector from A to B is $(d,\ -0.1061)$ m. Vertical components:

$$0 = a_{Ay} + \alpha_{AB}\,d + \omega_{AB}^2(0.1061) \text{ with } a_{Ay} = -42.43 \;\Rightarrow\; \alpha_{AB} = \frac{42.43 - \omega_{AB}^2(0.1061)}{d}$$ $$\alpha_{AB} = \frac{42.43 - (12.903)(0.1061)}{0.5906} = 69.52\ \text{rad/s}^2\ \text{(counter-clockwise)}$$

Horizontal direction:

$$a_B = a_{Ax} + \alpha_{AB}(0.1061) - \omega_{AB}^2\,d$$ $$a_B = -42.43 + (69.52)(0.1061) - (12.903)(0.5906) = -42.67\ \text{m/s}^2$$

The slider's acceleration is directed toward the crank centre O.

Check (approximate formula): $a_B \approx -\omega^2 r\left(\cos\theta + \dfrac{\cos2\theta}{n}\right)$ with $n = l/r = 4$ gives $-60(0.7071 + 0) = -42.4$ m/s², which agrees closely.

Answer: a_B = 42.67 m/s² toward O; α_AB = 69.52 rad/s² counter-clockwise.