Plane Motion of Rigid Bodies: Forces, Moments, and Accelerations

Take upward as positive for translation. The disk is homogeneous, so $I = \tfrac12 mr^2$. The forces on the disk are its weight $W=mg$ (down, at $G$) and the cord tension $T$ (up, at the rim, a distance $r$ from $G$). The cord leaves the rim tangentially, so $T$ has moment $Tr$ about $G$.

        T ^
          | A
      .---+---.
     /    G    \
     \         /
      '-------'
          | W
          v

Data

$m=15$ kg, $r=0.5$ m, $T=180$ N, $g = 9.81$ m/s²

$$W = mg = 15\times9.81 = 147.15\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(15)(0.5)^2 = 1.875\ \text{kg m}^2$$

Equations of motion

$$\begin{aligned} \Sigma F_y = m\bar a &: \quad T - W = m\bar a\\ \Sigma M_G = I\alpha &: \quad T r = I\alpha \end{aligned}$$

Centre of the disk

$$\bar a = \frac{T - W}{m} = \frac{180 - 147.15}{15} = 2.19\ \text{m/s}^2\ upward$$

Angular acceleration

$$\alpha = \frac{Tr}{I} = \frac{180\times0.5}{1.875} = 48.00\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}$$

Acceleration of the cord

The cord has the same acceleration as the rim point $A$ to which it is attached: the acceleration of $G$ plus the tangential acceleration $\alpha r$ of $A$ relative to $G$ (taken upward, since $\alpha$ is counter-clockwise and $A$ is on the right):

$$a_{cord} = \bar a_y + \alpha r = (+2.19) + 48.00\times0.5 = 26.19\ \text{m/s}^2\ \uparrow$$

Answer: (a) $\bar a = 2.19$ m/s² upward; (b) $\alpha = 48.00$ rad/s² counter-clockwise; (c) $a_{cord} = 26.19$ m/s² upward.

Take upward as positive for translation. The disk is homogeneous, so $I = \tfrac12 mr^2$. The forces on the disk are its weight $W=mg$ (down, at $G$) and the cord tension $T$ (up, at the rim, a distance $r$ from $G$). The cord leaves the rim tangentially, so $T$ has moment $Tr$ about $G$.

        T ^
          | A
      .---+---.
     /    G    \
     \         /
      '-------'
          | W
          v

Data

$m=15$ kg, $r=0.5$ m, $T=180$ N, $g = 9.81$ m/s²

$$W = mg = 15\times9.81 = 147.15\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(15)(0.5)^2 = 1.875\ \text{kg m}^2$$

Equations of motion

$$\begin{aligned} \Sigma F_y = m\bar a &: \quad T - W = m\bar a\\ \Sigma M_G = I\alpha &: \quad T r = I\alpha \end{aligned}$$

Angular acceleration

$$\alpha = \frac{Tr}{I} = \frac{180\times0.5}{1.875} = 48.00\ \text{rad/s}^2\ \text{(counter-clockwise)}$$

Acceleration of the cord

First, the centre: $\bar a = (T-W)/m = (180 - 147.15)/15 = 2.19$ m/s² upward. Cord acceleration = acceleration of rim point $A$:

$$a_{cord} = \bar a + \alpha r = 2.19 + 48.00\times0.5 = 26.19\ \text{m/s}^2\ \uparrow$$

Answer: (i) $\alpha = 48.00$ rad/s² counter-clockwise; (ii) $a_{cord} = 26.19$ m/s² upward.

Take upward as positive for translation. The disk is homogeneous, so $I = \tfrac12 mr^2$. The forces on the disk are its weight $W=mg$ (down, at $G$) and the pull $T$ (the given force) (up, at the rim, a distance $r$ from $G$). The cord leaves the rim tangentially, so $T$ has moment $Tr$ about $G$.

        T ^
          | A
      .---+---.
     /    G    \
     \         /
      '-------'
          | W
          v

Data

$m=20$ kg, $r=0.5$ m, $T=250$ N (the force called $F$ in the question), $g = 9.81$ m/s²

$$W = mg = 20\times9.81 = 196.20\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(20)(0.5)^2 = 2.5\ \text{kg m}^2$$

Equations of motion

$$\begin{aligned} \Sigma F_y = m\bar a &: \quad T - W = m\bar a\\ \Sigma M_G = I\alpha &: \quad T r = I\alpha \end{aligned}$$

Centre of the disk

$$\bar a = \frac{T - W}{m} = \frac{250 - 196.20}{20} = 2.69\ \text{m/s}^2\ upward$$

Angular acceleration

$$\alpha = \frac{Tr}{I} = \frac{250\times0.5}{2.5} = 50.00\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}$$

Acceleration of the cord

The cord has the same acceleration as the rim point $A$ to which it is attached: the acceleration of $G$ plus the tangential acceleration $\alpha r$ of $A$ relative to $G$ (taken upward, since $\alpha$ is counter-clockwise and $A$ is on the right):

$$a_{cord} = \bar a_y + \alpha r = (+2.69) + 50.00\times0.5 = 27.69\ \text{m/s}^2\ \uparrow$$

Answer: (a) angular acceleration $\alpha = 50.00$ rad/s² counter-clockwise; (b) acceleration of the disk $\bar a = 2.69$ m/s² upward; (c) acceleration of the cord $= 27.69$ m/s² upward.

Take upward as positive for translation. The disk is homogeneous, so $I = \tfrac12 mr^2$. The forces on the disk are its weight $W=mg$ (down, at $G$) and the pull $T$ (the given force) (up, at the rim, a distance $r$ from $G$). The cord leaves the rim tangentially, so $T$ has moment $Tr$ about $G$.

        T ^
          | A
      .---+---.
     /    G    \
     \         /
      '-------'
          | W
          v

Data

$m=30$ kg, $r=0.5$ m, $T=200$ N, $g = 9.81$ m/s²

$$W = mg = 30\times9.81 = 294.30\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(30)(0.5)^2 = 3.75\ \text{kg m}^2$$

Equations of motion

$$\begin{aligned} \Sigma F_y = m\bar a &: \quad T - W = m\bar a\\ \Sigma M_G = I\alpha &: \quad T r = I\alpha \end{aligned}$$

Centre of the disk

$$\bar a = \frac{T - W}{m} = \frac{200 - 294.30}{30} = 3.14\ \text{m/s}^2\ downward$$

Angular acceleration

$$\alpha = \frac{Tr}{I} = \frac{200\times0.5}{3.75} = 26.67\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}$$

Acceleration of the cord

The cord has the same acceleration as the rim point $A$ to which it is attached: the acceleration of $G$ plus the tangential acceleration $\alpha r$ of $A$ relative to $G$ (taken upward, since $\alpha$ is counter-clockwise and $A$ is on the right):

$$a_{cord} = \bar a_y + \alpha r = (-3.14) + 26.67\times0.5 = 10.19\ \text{m/s}^2\ \uparrow$$

Answer: (a) angular acceleration $\alpha = 26.67$ rad/s² counter-clockwise; (b) acceleration of the disk $\bar a = 3.14$ m/s² downward; (c) acceleration of the cord $= 10.19$ m/s² upward.

Principle of impulse and momentum for plane motion of a rigid body: the momenta of all particles of the body at time $t_1$, plus the impulses of the external forces acting between $t_1$ and $t_2$, equal the momenta at time $t_2$. The momenta of the particles reduce to a linear momentum vector $m\bar{\vec v}$ at the mass centre and an angular momentum $\bar I\omega$ about it:

$$\begin{aligned} m\bar{\vec v}_1 + \sum\int \vec F\,dt &= m\bar{\vec v}_2\\ \bar I\omega_1 + \sum\int M_G\,dt &= \bar I\omega_2 \end{aligned}$$

The moment equation may also be written about a fixed point $O$ (or any point with the appropriate momentum terms): $H_{O1} + \sum\int M_O\,dt = H_{O2}$. In scalar form the principle gives three equations (two for the linear part, one for the rotational part) and it is the best method when forces are given as a function of time or when the interval, not the distance, is of interest, as in impact problems.

 (momenta)1   +   (impulses)   =   (momenta)2

   m v1  .--.         F dt          m v2  .--.
   ---> (G) I w1  +   ------>  =    ---> (G) I w2
        '--'                             '--'

Numerical part

Take upward as positive for translation. The disk is homogeneous, so $I = \tfrac12 mr^2$. The forces on the disk are its weight $W=mg$ (down, at $G$) and the cord tension $T$ (up, at the rim, a distance $r$ from $G$). The cord leaves the rim tangentially, so $T$ has moment $Tr$ about $G$.

        T ^
          | A
      .---+---.
     /    G    \
     \         /
      '-------'
          | W
          v

Data

$m=15$ kg, $r=0.5$ m, $T=200$ N, $g = 9.81$ m/s²

$$W = mg = 15\times9.81 = 147.15\ \text{N}, \qquad I = \tfrac12 mr^2 = \tfrac12(15)(0.5)^2 = 1.875\ \text{kg m}^2$$

Equations of motion

$$\begin{aligned} \Sigma F_y = m\bar a &: \quad T - W = m\bar a\\ \Sigma M_G = I\alpha &: \quad T r = I\alpha \end{aligned}$$

Centre of the disk

$$\bar a = \frac{T - W}{m} = \frac{200 - 147.15}{15} = 3.52\ \text{m/s}^2\ upward$$

Angular acceleration

$$\alpha = \frac{Tr}{I} = \frac{200\times0.5}{1.875} = 53.33\ \text{rad/s}^2\ \text{(counter-clockwise, as } T \text{ acts at the right rim)}$$

Acceleration of the cord

The cord has the same acceleration as the rim point $A$ to which it is attached: the acceleration of $G$ plus the tangential acceleration $\alpha r$ of $A$ relative to $G$ (taken upward, since $\alpha$ is counter-clockwise and $A$ is on the right):

$$a_{cord} = \bar a_y + \alpha r = (+3.52) + 53.33\times0.5 = 30.19\ \text{m/s}^2\ \uparrow$$

Answer: (a) $\bar a = 3.52$ m/s² upward; (b) $\alpha = 53.33$ rad/s² counter-clockwise; (c) $a_{cord} = 30.19$ m/s² upward.

Assumptions. The cord leaves the bottom of the inner drum (radius $r=0.060$ m) and pulls horizontally to the right; the wheel (outer radius $R=0.100$ m) rests on rough ground. Take $x$ to the right and counter-clockwise rotation positive; the friction force $F$ acts at the contact point.

Data

$m = 50$ kg, $\bar k = 0.070$ m, $P = 200$ N, $\mu_s = 0.20$, $\mu_k = 0.15$

$$\bar I = m\bar k^2 = 50(0.070)^2 = 0.2450\ \text{kg m}^2, \qquad N = mg = 50\times9.81 = 490.50\ \text{N}$$

Step 1: assume rolling without slipping

Then $\bar a = \alpha R$ (the wheel rolls to the right, $\alpha$ counter-clockwise). Taking $F$ as friction acting to the left:

$$\begin{aligned} P - F &= m\bar a\\ Pr - FR &= \bar I\alpha \end{aligned}$$

Eliminating $F$ with $\bar a=\alpha R$:

$$\alpha = \frac{P(R-r)}{m(\bar k^2 + R^2)} = \frac{200(0.100-0.060)}{50(0.0049+0.0100)} = 10.74\ \text{rad/s}^2, \qquad \bar a = \alpha R = 1.07\ \text{m/s}^2$$

Friction required: $F = P - m\bar a = 200 - 50(1.07) = 146.31$ N.

Maximum available: $F_{max} = \mu_s N = 0.20\times490.50 = 98.10$ N.

Since $F = 146.31\ \text{N} > F_{max}$, the wheel slips, so the assumption is wrong and kinetic friction acts: $F = \mu_k N = 0.15\times490.50 = 73.58$ N.

Step 2: sliding (kinetic friction)

$$\begin{aligned} \bar a &= \frac{P - F_k}{m} = \frac{200 - 73.58}{50} = 2.53\ \text{m/s}^2\ (\to)\\ \alpha &= \frac{Pr - F_kR}{\bar I} = \frac{200(0.060) - 73.58(0.100)}{0.2450} = 18.95\ \text{rad/s}^2\ \text{(counter-clockwise)} \end{aligned}$$

The contact point moves forward relative to the ground ($\bar a + \alpha R>0$), which confirms that friction acts to the left.

Answer: $\bar a_G = 2.53$ m/s² to the right; $\alpha = 18.95$ rad/s² counter-clockwise.

Assumptions. The cord leaves the bottom of the inner drum (radius $r=0.060$ m) and pulls horizontally to the right; the wheel (outer radius $R=0.100$ m) rests on rough ground. Take $x$ to the right and counter-clockwise rotation positive; the friction force $F$ acts at the contact point.

Data

$m = 45$ kg, $\bar k = 0.070$ m, $P = 200$ N, $\mu_s = 0.20$, $\mu_k = 0.15$

$$\bar I = m\bar k^2 = 45(0.070)^2 = 0.2205\ \text{kg m}^2, \qquad N = mg = 45\times9.81 = 441.45\ \text{N}$$

Step 1: assume rolling without slipping

Then $\bar a = \alpha R$ (the wheel rolls to the right, $\alpha$ counter-clockwise). Taking $F$ as friction acting to the left:

$$\begin{aligned} P - F &= m\bar a\\ Pr - FR &= \bar I\alpha \end{aligned}$$

Eliminating $F$ with $\bar a=\alpha R$:

$$\alpha = \frac{P(R-r)}{m(\bar k^2 + R^2)} = \frac{200(0.100-0.060)}{45(0.0049+0.0100)} = 11.93\ \text{rad/s}^2, \qquad \bar a = \alpha R = 1.19\ \text{m/s}^2$$

Friction required: $F = P - m\bar a = 200 - 45(1.19) = 146.31$ N.

Maximum available: $F_{max} = \mu_s N = 0.20\times441.45 = 88.29$ N.

Since $F = 146.31\ \text{N} > F_{max}$, the wheel slips, so the assumption is wrong and kinetic friction acts: $F = \mu_k N = 0.15\times441.45 = 66.22$ N.

Step 2: sliding (kinetic friction)

$$\begin{aligned} \bar a &= \frac{P - F_k}{m} = \frac{200 - 66.22}{45} = 2.97\ \text{m/s}^2\ (\to)\\ \alpha &= \frac{Pr - F_kR}{\bar I} = \frac{200(0.060) - 66.22(0.100)}{0.2205} = 24.39\ \text{rad/s}^2\ \text{(counter-clockwise)} \end{aligned}$$

The contact point moves forward relative to the ground ($\bar a + \alpha R>0$), which confirms that friction acts to the left.

Answer: $\bar a_G = 2.97$ m/s² to the right; $\alpha = 24.39$ rad/s² counter-clockwise.

D'Alembert's principle converts a dynamics problem into a statics problem. A body in motion is in dynamic equilibrium if, in addition to the real forces and moments, we add inertia effects equal and opposite to the effective forces:

• Inertia force $-m\bar{\vec a}$ acting at the mass centre $G$.
• Inertia couple $-\bar I\vec\alpha$ (for plane motion).

For a rigid body in plane motion the equations of dynamic equilibrium are

$$\sum F_x - m\bar a_x = 0, \qquad \sum F_y - m\bar a_y = 0, \qquad \sum M_G - \bar I\alpha = 0$$

which are the same as $\sum \vec F = m\bar{\vec a}$ and $\sum M_G = \bar I\alpha$, but written as "sum of forces including inertia terms is zero". The moment equation can then be taken about any point, as in statics. Special cases: pure translation ($\alpha=0$): only the inertia force; rotation about a fixed axis through $G$: only the inertia couple $-I\alpha$.

   real forces + (-m a) at G + (-I alpha) couple  =>  equilibrium

Numerical part

Data: $m_A = 10$ kg, $\bar k_A = 0.080$ m, $r_A = 0.250$ m, $r_B = 0.100$ m, $M = 8$ N m on gear $B$. The data give no mass for gear $B$, so its inertia is neglected (if it is included, add $I_B$ in the equations below). The tangential force between the teeth is $F$.

$$\bar I_A = m_A\bar k_A^2 = 10(0.080)^2 = 0.064\ \text{kg m}^2$$

        M (on B)
         ( B  )---F--->(tooth)
                      ( A )

(b) Tangential force on A

Gear $B$ (massless): $M - F r_B = 0$

$$F = \frac{M}{r_B} = \frac{8}{0.100} = 80.00\ \text{N}$$

(a) Time to reach 600 rpm

Gear $A$: $F r_A = \bar I_A\alpha_A$

$$\alpha_A = \frac{F r_A}{\bar I_A} = \frac{80.00\times0.250}{0.064} = 312.50\ \text{rad/s}^2$$

The teeth have the same tangential acceleration: $\alpha_B r_B = \alpha_A r_A$

$$\alpha_B = \alpha_A\frac{r_A}{r_B} = 312.50\times\frac{0.250}{0.100} = 781.25\ \text{rad/s}^2$$ $$\omega_B = 600\ \text{rpm} = \frac{600\times2\pi}{60} = 62.83\ \text{rad/s}, \qquad t = \frac{\omega_B}{\alpha_B} = \frac{62.83}{781.25} = 0.0804\ \text{s}$$

Answer: $t = 0.0804$ s; tangential force $F = 80.00$ N.

Gear $E$ and the arm $OB$ rotate together about $O$, so the tangential force $F_D$ of gear $D$ at the pitch radius $r_E = 0.120$ m must supply the torque needed to give both bodies the angular acceleration $\alpha = 40$ rad/s². The angular velocity (8 rad/s) does not enter the moment equation, since both bodies rotate about their own fixed axis $O$ (centripetal terms produce no moment about $O$).

Assumptions. The mechanism turns in a horizontal plane (or $G$'s weight has no moment about $O$ at this instant). Arm $OB$ is taken as a uniform slender rod pivoted at its end $O$ with length $OB = 0.400$ m; the figure's 40 mm dimension is not used for the length. If the length differs, use $I_{OB} = \tfrac13 m L^2$ with the actual $L$.

Moments of inertia about O

$$\bar I_E = m_E\bar k_E^2 = 4(0.085)^2 = 0.0289\ \text{kg m}^2$$ $$I_{OB} = \tfrac13 m_{OB}L^2 = \tfrac13(3)(0.400)^2 = 0.160\ \text{kg m}^2$$ $$I_O = I_E + I_{OB} = 0.1889\ \text{kg m}^2$$

Equation of motion about O

$$F_D\,r_E = I_O\,\alpha$$ $$F_D = \frac{I_O\alpha}{r_E} = \frac{0.1889\times40}{0.120} = 62.97\ \text{N}$$

The moment is $M_O = I_O\alpha = 7.56$ N m.

Answer: tangential force of gear $D$ on gear $E$ is about $62.97$ N (acting in the sense of the counter-clockwise angular acceleration).

Angular momentum of a rigid body in plane motion is the sum of the moments of the momenta of all its particles about a point. For a body of mass $m$ in plane motion with angular velocity $\omega$, the momenta of the particles are equivalent to the linear momentum $m\bar{\vec v}$ at the mass centre $G$ together with a couple $\bar I\omega$.

About the mass centre

$$H_G = \bar I\,\omega$$

where $\bar I$ is the moment of inertia about the axis through $G$ perpendicular to the plane. This is true whether or not $G$ is moving.

About any other point O

$$H_O = \bar I\omega + (\bar{\vec r}\times m\bar{\vec v}) = \bar I\omega + m\bar v\,d$$

where $d$ is the perpendicular distance from $O$ to the line of action of $m\bar{\vec v}$.

Special cases

The rate of change of angular momentum equals the sum of moments: $\sum M_G = \dot H_G = \bar I\alpha$.

Examples

1. A flywheel of mass 20 kg and radius of gyration 0.3 m turning at 100 rad/s about its axis: $H = mk^2\omega = 20(0.3)^2(100) = 180$ kg m²/s.
2. A uniform disk (mass 10 kg, radius 0.5 m) rolling at $\bar v = 2$ m/s: $\omega = 4$ rad/s, $H_G = \tfrac12(10)(0.5)^2(4) = 5$ kg m²/s; about the contact point $H_C = 5 + 10(2)(0.5) = 15$ kg m²/s, which equals $I_C\omega = \tfrac32(10)(0.25)(4)= 15$ kg m²/s.