Kinematics of Rigid Bodies

The lower rack is stationary and the gear rolls on it without slipping, so the contact point $C$ is the instantaneous centre. Take $x$ to the right, $y$ up. $B$ is on the inner radius ($r_2=0.100$ m) above $A$; $C$ is at $r_1 = 0.150$ m below $A$; $D$ is taken on the outer radius, $0.150$ m to the left of $A$.

(a) Angular acceleration

$$\omega = \frac{v_A}{r_1} = \frac{1.2}{0.150} = 8.00\ \text{rad/s (clockwise)}, \qquad \alpha = \frac{a_A}{r_1} = \frac{3}{0.150} = 20.00\ \text{rad/s}^2\ \text{(clockwise)}$$

(b) Accelerations of B, C and D

Use $\vec a_P = \vec a_A + \vec\alpha\times\vec r_{P/A} - \omega^2\vec r_{P/A}$ with $\vec\alpha = -20.00\,\hat k$, $\omega^2 = 64.00$ rad²/s², $\vec a_A = 3\,\hat i$.

Point B, $\vec r_{B/A} = 0.100\,\hat j$:

$$\vec a_B = 3\hat i + (-20.00\hat k)\times(0.1\hat j) - 64.00(0.1\hat j) = (5.00\,\hat i -6.40\,\hat j)\ \text{m/s}^2$$

$a_B = 8.12$ m/s² at 52.0° below the $x$-axis ($\theta$ measured towards $+x$).

Point C, $\vec r_{C/A} = -0.150\,\hat j$:

$$\vec a_C = 3\hat i + (-20.00\hat k)\times(-0.15\hat j) - 64.00(-0.15\hat j) = (0.00\,\hat i +9.60\,\hat j)\ \text{m/s}^2$$

$a_C = 9.60$ m/s² directed upward (towards $A$, the centripetal acceleration of the contact point; its horizontal part is zero).

Point D, $\vec r_{D/A} = -0.150\,\hat i$:

$$\vec a_D = 3\hat i + (-20.00\hat k)\times(-0.15\hat i) - 64.00(-0.15\hat i) = (12.60\,\hat i +3.00\,\hat j)\ \text{m/s}^2$$

$a_D = 12.95$ m/s² at 13.4° above the $x$-axis.

Answer: $\alpha = 20.00$ rad/s² clockwise; $a_B = 8.12$ m/s², $a_C = 9.60$ m/s² (upward), $a_D = 12.95$ m/s².

Coriolis acceleration is the extra acceleration a point has when it moves relative to a rotating (moving) frame or body. If a point $P$ moves with relative velocity $\vec v_{rel}$ on a body rotating with angular velocity $\vec\omega$, then

$$\vec a_P = \vec a_{P'} + \vec a_{rel} + 2\vec\omega\times\vec v_{rel}$$

The velocity problem below involves only points of rigid links joined by pins, so no Coriolis term appears.

Assumed layout (from the figure): $D$ and $A$ are fixed pins at the same level; $CD$ is vertical (0.6 m) with $C$ at the top; $BC$ is horizontal (0.6 m) with $B$ to the left of $C$ and $E$ its midpoint; $AB$ rises at $30^\circ$ from $A$ to $B$, so $AB = 0.6/\sin30^\circ = 1.2$ m. Take $\omega_{CD}=6$ rad/s counter-clockwise, $x$ to the right and $y$ up (reverse all senses if $CD$ turns the other way).

Step 1: velocity of C

$$\vec v_C = \omega_{CD}\hat k\times\vec r_{C/D} = 6\hat k\times0.6\hat j = -3.6\,\hat i\ \text{m/s}$$

Step 2: direction of v_B

Link $AB$ rotates about $A$, so $\vec v_B \perp AB$. With $\vec v_B = (v_{Bx}, v_{By})$ and $AB$ along $(\cos30^\circ, \sin30^\circ)$:

$$v_{Bx}\cos30^\circ + v_{By}\sin30^\circ = 0$$

Link $BC$ is horizontal, so $\omega_{BC}\hat k\times\vec r_{B/C}$ is vertical and $v_{Bx} = v_{Cx} = -3.6$ m/s. Hence

$$v_{By} = \frac{3.6\cos30^\circ}{\sin30^\circ} = 6.24\ \text{m/s}, \qquad v_B = \sqrt{3.6^2 + 6.24^2} = 7.20\ \text{m/s}$$

Step 3: link BC

$$\vec v_B = \vec v_C + \omega_{BC}\hat k\times(-0.6\hat i) \;\Rightarrow\; v_{By} = -0.6\,\omega_{BC} \;\Rightarrow\; \omega_{BC} = -10.39\ \text{rad/s}\ \text{(clockwise)}$$

Step 4: point E

$$\vec v_E = \vec v_C + \omega_{BC}\hat k\times(-0.3\hat i) = -3.6\hat i + (-10.39)(-0.3)\hat j = (-3.60\,\hat i + 3.12\,\hat j)\ \text{m/s}$$

$v_E = 4.76$ m/s, directed 40.9° above the horizontal, towards the left.

Answer: $v_E = 4.76$ m/s (40.9° above horizontal, towards the left); $\omega_{BC} = 10.39$ rad/s clockwise.

Assumed layout (from the figure): $D$ and $A$ are fixed pins at the same level; $CD$ is vertical (0.6 m) with $C$ at the top; $BC$ is horizontal (0.6 m) with $B$ to the left of $C$ and $E$ its midpoint; $AB$ rises at $30^\circ$ from $A$ to $B$, so $AB = 0.6/\sin30^\circ = 1.2$ m. Take $\omega_{CD}=6$ rad/s counter-clockwise, $x$ to the right and $y$ up (reverse all senses if $CD$ turns the other way).

Step 1: velocity of C

$$\vec v_C = \omega_{CD}\hat k\times\vec r_{C/D} = 6\hat k\times0.6\hat j = -3.6\,\hat i\ \text{m/s}$$

Step 2: direction of v_B

Link $AB$ rotates about $A$, so $\vec v_B \perp AB$. With $\vec v_B = (v_{Bx}, v_{By})$ and $AB$ along $(\cos30^\circ, \sin30^\circ)$:

$$v_{Bx}\cos30^\circ + v_{By}\sin30^\circ = 0$$

Link $BC$ is horizontal, so $\omega_{BC}\hat k\times\vec r_{B/C}$ is vertical and $v_{Bx} = v_{Cx} = -3.6$ m/s. Hence

$$v_{By} = \frac{3.6\cos30^\circ}{\sin30^\circ} = 6.24\ \text{m/s}, \qquad v_B = \sqrt{3.6^2 + 6.24^2} = 7.20\ \text{m/s}$$

Step 3: link BC

$$\vec v_B = \vec v_C + \omega_{BC}\hat k\times(-0.6\hat i) \;\Rightarrow\; v_{By} = -0.6\,\omega_{BC} \;\Rightarrow\; \omega_{BC} = -10.39\ \text{rad/s}\ \text{(clockwise)}$$

Step 4: point E

$$\vec v_E = \vec v_C + \omega_{BC}\hat k\times(-0.3\hat i) = -3.6\hat i + (-10.39)(-0.3)\hat j = (-3.60\,\hat i + 3.12\,\hat j)\ \text{m/s}$$

$v_E = 4.76$ m/s, directed 40.9° above the horizontal, towards the left.

Step 5: angular velocity of AB

$$\omega_{AB} = \frac{v_B}{AB} = \frac{7.20}{1.2} = 6.00\ \text{rad/s}\ \text{(counter-clockwise)}$$

Answer: $v_E = 4.76$ m/s; $\omega_{AB} = 6.00$ rad/s counter-clockwise.

General plane motion

General plane motion (GPM) is the motion of a rigid body in which every particle stays in a plane parallel to a fixed plane, and the motion is neither pure translation nor pure rotation about a fixed axis. It is the sum of (i) a translation of a chosen reference point and (ii) a rotation of the body about that point.

For two points $A$ and $B$ of the same rigid body:

$$\vec v_B = \vec v_A + \vec\omega \times \vec r_{B/A}, \qquad \vec a_B = \vec a_A + \vec\alpha \times \vec r_{B/A} - \omega^2 \vec r_{B/A}$$

Here $v_{B/A} = \omega\, r_{B/A}$ is perpendicular to $AB$, and $a_{B/A}$ has a tangential part $\alpha r$ and a normal part $\omega^2 r$ towards $A$.

Examples

• A wheel rolling on the road: the centre translates while the wheel rotates about the centre.
• The connecting rod of an engine: end $A$ (crank pin) moves on a circle, end $B$ (piston) moves in a straight line.
• A ladder sliding down a wall.

Instantaneous centre of rotation (ICR) is the point in the plane of motion (on the body or on its imaginary extension) whose velocity is zero at that instant. At that instant the body behaves as if it were rotating about this point, so every point has speed $v = \omega\,r$ with $r$ measured from the ICR, perpendicular to the line joining the point to the ICR.

Geometry

Take $A$ as origin, $x$ horizontal towards the piston, $y$ upward. The crank turns clockwise, so $\omega_{AB} = -188.50$ rad/s ($1800$ rpm $= 1800\times 2\pi/60$). The piston $D$ moves only horizontally, so $y_D = 0$:

$$\sin\beta = \frac{AB\sin 40^\circ}{BD} = \frac{3\sin 40^\circ}{8} = 0.2410 \;\Rightarrow\; \beta = 13.95^\circ$$ $$\vec r_{B/A} = (2.2981\,\hat i + 1.9284\,\hat j)\ \text{in}, \qquad \vec r_{D/B} = (7.7641\,\hat i - 1.9284\,\hat j)\ \text{in}$$

Velocities

Crank pin $B$ moves on a circle about $A$:

$$\vec v_B = \vec\omega_{AB}\times\vec r_{B/A} = (363.49\,\hat i -433.19\,\hat j)\ \text{in/s}, \qquad v_B = \omega_{AB}\,AB = 565.49\ \text{in/s}$$

For the connecting rod, $\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B}$ with $\vec v_D = v_D\hat i$:

$$\begin{aligned} \hat j:&\quad 0 = -433.19 + \omega_{BD}(7.7641) \;\Rightarrow\; \omega_{BD} = 55.79\ \text{rad/s}\\ \hat i:&\quad v_D = 363.49 + \omega_{BD}(1.9284) = 471.08\ \text{in/s} \end{aligned}$$

Accelerations

The crank speed is constant ($\alpha_{AB}=0$), so $B$ has only a normal acceleration towards $A$:

$$\vec a_B = -\omega_{AB}^2\,\vec r_{B/A} = (-81654.0\,\hat i -68515.8\,\hat j)\ \text{in/s}^2 \quad (a_B = 106591.7\ \text{in/s}^2)$$

For the rod, with $\omega_{BD} = 55.79$ rad/s (counter-clockwise) and $\vec a_D = a_D\hat i$:

$$\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\,\vec r_{D/B}$$ $$\omega_{BD}^2\,\vec r_{D/B} = (24169.1\,\hat i -6002.9\,\hat j)\ \text{in/s}^2, \qquad \hat k\times\vec r_{D/B} = (1.9284\,\hat i + 7.7641\,\hat j)\ \text{in}$$

Equating components:

$$\begin{aligned} \hat j:&\quad 0 = -68515.8 +6002.9 + \alpha_{BD}(7.7641) \;\Rightarrow\; \alpha_{BD} = 8051.5\ \text{rad/s}^2\\ \hat i:&\quad a_D = -81654.0 -24169.1 + \alpha_{BD}(1.9284) = -90296.9\ \text{in/s}^2 \end{aligned}$$

Answer: $\alpha_{BD} \approx 8051.5$ rad/s² (counter-clockwise), $a_D \approx 90296.9$ in/s² (towards the crank, i.e. to the left).

Answer: $\omega_{BD} \approx 55.79$ rad/s counter-clockwise; $v_P = v_D \approx 471.08$ in/s $= 39.26$ ft/s (to the right).

Instantaneous centre of rotation (ICR) is the point in the plane of motion (on the body or on its imaginary extension) whose velocity is zero at that instant. At that instant the body behaves as if it were rotating about this point, so every point has speed $v = \omega\,r$ with $r$ measured from the ICR, perpendicular to the line joining the point to the ICR.

Location: draw perpendiculars to the velocity directions of two points of the body; they meet at the ICR. For a body rolling without slipping, the ICR is the contact point.

Examples

• Rolling wheel: ICR is the point of contact with the ground, so $v_{centre} = \omega r$ and the top point moves at $2\omega r$.
• Ladder sliding on a wall and floor: ICR is the intersection of the vertical through the foot and the horizontal through the top end.
• Connecting rod of an engine: ICR lies where the perpendicular to the piston path through the piston meets the perpendicular to the crank-pin velocity.

Acceleration analysis

The values $\omega_{BD}=61.87$ rad/s (ccw) and $v_D = 13.26$ m/s are given and are used directly.

Geometry

Take $A$ as origin, $x$ horizontal towards the piston, $y$ upward. The crank turns clockwise, so $\omega_{AB} = -209.44$ rad/s ($2000$ rpm $= 2000\times 2\pi/60$). The piston $D$ moves only horizontally, so $y_D = 0$:

$$\sin\beta = \frac{AB\sin 40^\circ}{BD} = \frac{0.076\sin 40^\circ}{0.203} = 0.2406 \;\Rightarrow\; \beta = 13.92^\circ$$ $$\vec r_{B/A} = (0.0582\,\hat i + 0.0489\,\hat j)\ \text{m}, \qquad \vec r_{D/B} = (0.1970\,\hat i - 0.0489\,\hat j)\ \text{m}$$

Accelerations

The crank speed is constant ($\alpha_{AB}=0$), so $B$ has only a normal acceleration towards $A$:

$$\vec a_B = -\omega_{AB}^2\,\vec r_{B/A} = (-2553.8\,\hat i -2142.9\,\hat j)\ \text{m/s}^2 \quad (a_B = 3333.7\ \text{m/s}^2)$$

For the rod, with $\omega_{BD} = 61.87$ rad/s (counter-clockwise) and $\vec a_D = a_D\hat i$:

$$\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\,\vec r_{D/B}$$ $$\omega_{BD}^2\,\vec r_{D/B} = (754.2\,\hat i -187.0\,\hat j)\ \text{m/s}^2, \qquad \hat k\times\vec r_{D/B} = (0.0489\,\hat i + 0.1970\,\hat j)\ \text{m}$$

Equating components:

$$\begin{aligned} \hat j:&\quad 0 = -2142.9 +187.0 + \alpha_{BD}(0.1970) \;\Rightarrow\; \alpha_{BD} = 9926.6\ \text{rad/s}^2\\ \hat i:&\quad a_D = -2553.8 -754.2 + \alpha_{BD}(0.0489) = -2823.1\ \text{m/s}^2 \end{aligned}$$

Answer: $\alpha_{BD} \approx 9926.6$ rad/s² (counter-clockwise), $a_D \approx 2823.1$ m/s² (towards the crank, i.e. to the left).

Coriolis acceleration is the extra acceleration a point has when it moves relative to a rotating (moving) frame or body. If a point $P$ moves with relative velocity $\vec v_{rel}$ on a body rotating with angular velocity $\vec\omega$, then

$$\vec a_P = \vec a_{P'} + \vec a_{rel} + 2\vec\omega\times\vec v_{rel}$$

where $\vec a_{P'}$ is the acceleration of the coincident point of the rotating frame. The term $\vec a_c = 2\vec\omega\times\vec v_{rel}$ is the Coriolis acceleration. Its magnitude is $2\omega v_{rel}$ and its direction is that of $\vec v_{rel}$ rotated through $90^\circ$ in the sense of $\omega$. It exists only when the point slides on the rotating body (for example a collar sliding on a rotating arm, or a slider in a slotted rotating link); it is zero for two points fixed in the same rigid body, such as the crank pin and piston pin of a connecting rod.

In the engine below, points $B$ and $D$ both belong to the rigid rod $BD$, so there is no relative sliding and the Coriolis term is zero; the relative acceleration is only $\alpha r$ and $\omega^2 r$.

Geometry

Use $\beta=13.9^\circ$ and $\omega_{BD}=61.9$ rad/s (ccw) as given.

Take $A$ as origin, $x$ horizontal towards the piston, $y$ upward. The crank turns clockwise, so $\omega_{AB} = -209.44$ rad/s ($2000$ rpm $= 2000\times 2\pi/60$). The piston $D$ moves only horizontally, so $y_D = 0$:

$$\sin\beta = \frac{AB\sin 40^\circ}{BD} = \frac{0.075\sin 40^\circ}{0.2} = 0.2410 \;\Rightarrow\; \beta = 13.95^\circ$$ $$\vec r_{B/A} = (0.0575\,\hat i + 0.0482\,\hat j)\ \text{m}, \qquad \vec r_{D/B} = (0.1941\,\hat i - 0.0482\,\hat j)\ \text{m}$$

Accelerations

The crank speed is constant ($\alpha_{AB}=0$), so $B$ has only a normal acceleration towards $A$:

$$\vec a_B = -\omega_{AB}^2\,\vec r_{B/A} = (-2520.2\,\hat i -2114.7\,\hat j)\ \text{m/s}^2 \quad (a_B = 3289.9\ \text{m/s}^2)$$

For the rod, with $\omega_{BD} = 61.90$ rad/s (counter-clockwise) and $\vec a_D = a_D\hat i$:

$$\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\,\vec r_{D/B}$$ $$\omega_{BD}^2\,\vec r_{D/B} = (743.7\,\hat i -184.7\,\hat j)\ \text{m/s}^2, \qquad \hat k\times\vec r_{D/B} = (0.0482\,\hat i + 0.1941\,\hat j)\ \text{m}$$

Equating components:

$$\begin{aligned} \hat j:&\quad 0 = -2114.7 +184.7 + \alpha_{BD}(0.1941) \;\Rightarrow\; \alpha_{BD} = 9943.0\ \text{rad/s}^2\\ \hat i:&\quad a_D = -2520.2 -743.7 + \alpha_{BD}(0.0482) = -2784.6\ \text{m/s}^2 \end{aligned}$$

Answer: $\alpha_{BD} \approx 9943.0$ rad/s² (counter-clockwise), $a_D \approx 2784.6$ m/s² (towards the crank, i.e. to the left).

Geometry

Take $A$ as origin, $x$ horizontal towards the piston, $y$ upward. The crank turns clockwise, so $\omega_{AB} = -20.94$ rad/s ($200$ rpm $= 200\times 2\pi/60$). The piston $D$ moves only horizontally, so $y_D = 0$:

$$\sin\beta = \frac{AB\sin 60^\circ}{BD} = \frac{0.15\sin 60^\circ}{0.5} = 0.2598 \;\Rightarrow\; \beta = 15.06^\circ$$ $$\vec r_{B/A} = (0.0750\,\hat i + 0.1299\,\hat j)\ \text{m}, \qquad \vec r_{D/B} = (0.4828\,\hat i - 0.1299\,\hat j)\ \text{m}$$

Velocities

Crank pin $B$ moves on a circle about $A$:

$$\vec v_B = \vec\omega_{AB}\times\vec r_{B/A} = (2.72\,\hat i -1.57\,\hat j)\ \text{m/s}, \qquad v_B = \omega_{AB}\,AB = 3.14\ \text{m/s}$$

For the connecting rod, $\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B}$ with $\vec v_D = v_D\hat i$:

$$\begin{aligned} \hat j:&\quad 0 = -1.57 + \omega_{BD}(0.4828) \;\Rightarrow\; \omega_{BD} = 3.25\ \text{rad/s}\\ \hat i:&\quad v_D = 2.72 + \omega_{BD}(0.1299) = 3.14\ \text{m/s} \end{aligned}$$

Accelerations

The crank speed is constant ($\alpha_{AB}=0$), so $B$ has only a normal acceleration towards $A$:

$$\vec a_B = -\omega_{AB}^2\,\vec r_{B/A} = (-32.9\,\hat i -57.0\,\hat j)\ \text{m/s}^2 \quad (a_B = 65.8\ \text{m/s}^2)$$

For the rod, with $\omega_{BD} = 3.25$ rad/s (counter-clockwise) and $\vec a_D = a_D\hat i$:

$$\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\,\vec r_{D/B}$$ $$\omega_{BD}^2\,\vec r_{D/B} = (5.1\,\hat i -1.4\,\hat j)\ \text{m/s}^2, \qquad \hat k\times\vec r_{D/B} = (0.1299\,\hat i + 0.4828\,\hat j)\ \text{m}$$

Equating components:

$$\begin{aligned} \hat j:&\quad 0 = -57.0 +1.4 + \alpha_{BD}(0.4828) \;\Rightarrow\; \alpha_{BD} = 115.2\ \text{rad/s}^2\\ \hat i:&\quad a_D = -32.9 -5.1 + \alpha_{BD}(0.1299) = -23.0\ \text{m/s}^2 \end{aligned}$$

Answer: $\alpha_{BD} \approx 115.2$ rad/s² (counter-clockwise), $a_D \approx 23.0$ m/s² (towards the crank, i.e. to the left).

Assume the crank $AB$ turns counter-clockwise at $\omega_{AB}=5$ rad/s, with $A$ as origin, $B$ above and to the right of $A$, and the slider line $0.15$ m above $A$ with $C$ to the right of $B$ (the answer magnitudes are the same for the other sense of rotation).

Geometry

$$\vec r_{B/A} = 0.6(\cos30^\circ\hat i + \sin30^\circ\hat j) = (0.5196\,\hat i + 0.3000\,\hat j)\ \text{m}$$

$B$ is $0.300 - 0.150 = 0.150$ m above the slider line, so $\sin\phi = 0.150/0.300 = 0.5$ and $\phi=30^\circ$ (link $BC$ slopes down to the right):

$$\vec r_{C/B} = 0.3(\cos30^\circ\hat i - \sin30^\circ\hat j) = (0.2598\,\hat i - 0.1500\,\hat j)\ \text{m}$$

Velocities

$$\vec v_B = \omega_{AB}\hat k\times\vec r_{B/A} = 5\hat k\times(0.5196\hat i + 0.3000\hat j) = (-1.500\,\hat i + 2.598\,\hat j)\ \text{m/s}$$ $$\vec v_C = \vec v_B + \omega_{BC}\hat k\times\vec r_{C/B} = \vec v_B + \omega_{BC}(0.1500\,\hat i + 0.2598\,\hat j)$$

The slider moves horizontally, so the $\hat j$ component of $\vec v_C$ is zero:

$$0 = 2.598 + 0.2598\,\omega_{BC} \;\Rightarrow\; \omega_{BC} = -10.00\ \text{rad/s}\ \text{(clockwise)}$$ $$v_C = -1.500 + 0.1500(-10.00) = -3.00\ \text{m/s}$$

Answer: $v_C = 3.00$ m/s (horizontal, towards $A$), $\omega_{BC} = 10.00$ rad/s clockwise.

Coriolis acceleration

Coriolis acceleration is the extra acceleration a point has when it moves relative to a rotating (moving) frame or body. If a point $P$ moves with relative velocity $\vec v_{rel}$ on a body rotating with angular velocity $\vec\omega$, then

$$\vec a_P = \vec a_{P'} + \vec a_{rel} + 2\vec\omega\times\vec v_{rel}$$

where $\vec a_{P'}$ is the acceleration of the coincident point of the rotating frame. The term $\vec a_c = 2\vec\omega\times\vec v_{rel}$ is the Coriolis acceleration. Its magnitude is $2\omega v_{rel}$ and its direction is that of $\vec v_{rel}$ rotated through $90^\circ$ in the sense of $\omega$. It exists only when the point slides on the rotating body (for example a collar sliding on a rotating arm, or a slider in a slotted rotating link); it is zero for two points fixed in the same rigid body, such as the crank pin and piston pin of a connecting rod.

Assumed layout: $OA$ is vertical, so $v_A = \omega_{OA}\cdot OA$ is horizontal. Rod $AB$ makes $30^\circ$ with the horizontal, and $B$ slides in a vertical guide, so $v_B$ is vertical. Use the instantaneous centre $I$ of rod $AB$.

Locating the ICR

$I$ lies on the perpendicular to $v_A$ through $A$ (the vertical line through $A$) and on the perpendicular to $v_B$ through $B$ (the horizontal line through $B$):

   A *   --> vA (horizontal)
     |\
     | \  AB
   y |  \
     |   \
   I *----* B   (vB vertical)
        x

Horizontal and vertical distances of $B$ from $A$:

$$x = AB\cos30^\circ = 0.6\cos30^\circ = 0.5196\ \text{m}, \qquad y = AB\sin30^\circ = 0.6\sin30^\circ = 0.300\ \text{m}$$

So $IA = y = 0.300$ m and $IB = x = 0.5196$ m.

Velocities

$$v_A = \omega_{OA}\,OA = 12\times0.3 = 3.60\ \text{m/s}$$ $$\omega_{AB} = \frac{v_A}{IA} = \frac{3.60}{0.300} = 12.00\ \text{rad/s}$$ $$v_B = \omega_{AB}\,(IB) = 12.00\times 0.5196 = 6.24\ \text{m/s (vertical)}$$

The midpoint $M$ of $AB$ is at $IM = \sqrt{(x/2)^2 + (y/2)^2} = \sqrt{(0.2598)^2 + (0.150)^2} = 0.300$ m, so

$$v_M = \omega_{AB}\,(IM) = 12.00\times 0.300 = 3.60\ \text{m/s}$$

(perpendicular to $IM$).

Answer: $v_B = 6.24$ m/s, $v_M = 3.60$ m/s, $\omega_{AB} = 12.00$ rad/s.

Constrained motion is motion in which the path or position of a body is restricted by guides, pins, links or contact, so its coordinates are related by geometric equations and it cannot move freely in all directions. The restricted motion lets us find unknown velocities and accelerations from the geometry.

Examples

• A piston in a cylinder: it can move only along the cylinder axis.
• A pendulum bob on a rod: it is forced to move on a circular arc.
• A collar sliding on a fixed rod, or a block on a slotted guide.
• A wheel rolling without slipping: the centre speed and angular speed are related by $v = \omega r$.

Types of rigid body motion

A rigid body in plane motion can have:

In space, a body can also have rotation about a fixed point (a spinning top) and general three-dimensional motion.

Double gear: angular velocity

The lower rack is stationary, so the contact point $C$ ($r_1=0.150$ m below $A$) is the instantaneous centre:

$$\omega = \frac{v_A}{r_1} = \frac{1.2}{0.150} = 8.00\ \text{rad/s (clockwise)}$$

Velocities of R and D

The upper rack is in contact with the gear at $B$, $r_2 = 0.100$ m above $A$, and moves with point $B$. Distance from $C$ to $B$ is $0.150 + 0.100 = 0.250$ m:

$$v_R = v_B = \omega\,(CB) = 8.00\times0.250 = 2.00\ \text{m/s}\ \to$$

Point $D$ is on the horizontal line through $A$, $0.150$ m to the left of $A$ (on the outer circle). Its distance from $C$ is $CD = \sqrt{0.150^2 + 0.150^2} = 0.2121$ m:

$$v_D = \omega\,(CD) = 8.00\times0.2121 = 1.70\ \text{m/s}$$

Components: $\vec v_D = \vec v_A + \omega\hat k\times\vec r_{D/A}$ with $\omega = -8\hat k$: $\vec v_D = (1.2\,\hat i + 1.2\,\hat j)$ m/s, i.e. 45° above the horizontal towards the right.

Answer: $\omega = 8.00$ rad/s clockwise; $v_R = 2.00$ m/s to the right; $v_D = 1.70$ m/s at $45^\circ$ above horizontal.

General plane motion (GPM) is the motion of a rigid body in which every particle stays in a plane parallel to a fixed plane, and the motion is neither pure translation nor pure rotation about a fixed axis. It is the sum of (i) a translation of a chosen reference point and (ii) a rotation of the body about that point.

For two points $A$ and $B$ of the same rigid body:

$$\vec v_B = \vec v_A + \vec\omega \times \vec r_{B/A}, \qquad \vec a_B = \vec a_A + \vec\alpha \times \vec r_{B/A} - \omega^2 \vec r_{B/A}$$

Here $v_{B/A} = \omega\, r_{B/A}$ is perpendicular to $AB$, and $a_{B/A}$ has a tangential part $\alpha r$ and a normal part $\omega^2 r$ towards $A$.

Examples

• A wheel rolling on the road: the centre translates while the wheel rotates about the centre.
• The connecting rod of an engine: end $A$ (crank pin) moves on a circle, end $B$ (piston) moves in a straight line.
• A ladder sliding down a wall.

Given data and assumed geometry

Place $A$ at the origin with $AB$ horizontal (40 mm), $BD$ horizontal (50 mm) and arm $DE$ pivoted at $E$, which is 50 mm to the right of $D$ and 30 mm above it (the figure's dimensions 40, 50, 50 mm and 30 mm):

$$\vec r_{B/A} = 0.040\hat i,\quad \vec r_{D/B} = 0.050\hat i,\quad \vec r_{D/E} = (-0.050\,\hat i - 0.030\,\hat j)\ \text{m}$$

$\omega_{AB} = 15$ rad/s ccw, $\alpha_{AB}=0$. A and E are fixed.

Velocities

$$\vec v_B = 15\hat k\times0.04\hat i = 0.6\,\hat j\ \text{m/s}$$ $$\vec v_D = \vec v_B + \omega_{BD}\hat k\times\vec r_{D/B} = \omega_{DE}\hat k\times\vec r_{D/E}$$ $$\begin{aligned} \hat i:&\quad 0 = 0.03\,\omega_{DE} \;\Rightarrow\; \omega_{DE} = 0\ \text{rad/s}\\ \hat j:&\quad 0.6 + 0.05\,\omega_{BD} = -0.05\,\omega_{DE} \;\Rightarrow\; \omega_{BD} = -12.00\ \text{rad/s} \end{aligned}$$

Accelerations of B and D

$$\vec a_B = -\omega_{AB}^2\vec r_{B/A} = -225(0.04)\hat i = -9\,\hat i\ \text{m/s}^2$$

From rod $BD$: $\vec a_D = \vec a_B + \alpha_{BD}\hat k\times\vec r_{D/B} - \omega_{BD}^2\vec r_{D/B} = (-9 - 144(0.05))\hat i + 0.05\,\alpha_{BD}\hat j = -16.2\,\hat i + 0.05\,\alpha_{BD}\hat j$

From arm $DE$ (since $\omega_{DE}=0$): $\vec a_D = \alpha_{DE}\hat k\times\vec r_{D/E} = \alpha_{DE}(0.03\,\hat i - 0.05\,\hat j)$

Equating components:

$$\begin{aligned} \hat i:&\quad -16.2 = 0.03\,\alpha_{DE} \;\Rightarrow\; \alpha_{DE} = -540.0\ \text{rad/s}^2\\ \hat j:&\quad 0.05\,\alpha_{BD} = -0.05\,\alpha_{DE} \;\Rightarrow\; \alpha_{BD} = 540.0\ \text{rad/s}^2 \end{aligned}$$ $$\vec a_D = (-16.2\,\hat i + 27.0\,\hat j)\ \text{m/s}^2, \qquad a_D = 31.5\ \text{m/s}^2$$

Answer: $\alpha_{DE} = 540.0$ rad/s² clockwise; $a_D = 31.5$ m/s² (components $-16.2\,\hat i + 27.0\,\hat j$). The result depends on the assumed layout; the method is the same for the exact figure.