Plane motion of rigid bodies: energy and momentum methods

Angular momentum of the bar and bullet about the pivot $C$ is conserved during the impact (the reaction at $C$ has no moment about $C$; the weights are non-impulsive). Take the bullet mass $m_b = 0.040$ kg, $v_0 = 600$ m/s, bar mass $M = 7$ kg, $L = 0.600$ m, $h = 0.240$ m (distance of the pivot $C$ below $A$); the bullet enters at the lower end $B$.

     A  o
        |   G = centre of bar
   h    |
     C  o-- pivot
        |
     B  o<--- bullet v0

Distances from C

• Centre of bar $G$ from $C$: $d_G = L/2 - h = 0.300 - 0.240 = 0.060$ m (below $C$)
• Bullet point $B$ from $C$: $d_B = L - h = 0.600 - 0.240 = 0.360$ m

Moment of inertia of the bar about C

$$\bar I = \frac{1}{12}ML^2 = \frac{1}{12}(7)(0.6)^2 = 0.210\ \text{kg m}^2$$ $$I_C = \bar I + Md_G^2 = 0.210 + 7(0.060)^2 = 0.2352\ \text{kg m}^2$$

(a) Angular velocity just after impact

The bullet is embedded, so its speed after impact is $\omega d_B$. Conservation of angular momentum about $C$:

$$m_bv_0d_B = I_C\omega + m_b(\omega d_B)d_B$$ $$\omega = \frac{m_bv_0d_B}{I_C + m_bd_B^2} = \frac{0.04\times600\times0.360}{0.2352 + 0.04(0.360)^2} = 35.94\ \text{rad/s}$$

(b) Impulsive reaction at C

Take the bullet direction as positive. Linear impulse-momentum for the system (bar + bullet), the impulse of the reaction being $C\,\Delta t$:

$$m_bv_0 + C\,\Delta t = M(\omega d_G) + m_b(\omega d_B)$$ $$C\,\Delta t = 7(35.94)(0.060) + 0.04(35.94)(0.360) - 0.04(600) = -8.39\ \text{N s}$$ $$C = \frac{C\,\Delta t}{\Delta t} = \frac{-8.39}{0.001} = -8387\ \text{N}$$

The negative sign means the reaction at $C$ acts opposite to the bullet's direction.

Answer: (a) $\omega = 35.94$ rad/s; (b) $C \approx 8387$ N ($\approx 8.39$ kN) opposite to the bullet velocity.

Angular momentum of the bar and bullet about the pivot $C$ is conserved during the impact (the reaction at $C$ has no moment about $C$; the weights are non-impulsive). Take the bullet mass $m_b = 0.040$ kg, $v_0 = 600$ m/s, bar mass $M = 7$ kg, $L = 0.600$ m, $h = 0.260$ m (distance of the pivot $C$ below $A$); the bullet enters at the lower end $B$.

     A  o
        |   G = centre of bar
   h    |
     C  o-- pivot
        |
     B  o<--- bullet v0

Distances from C

• Centre of bar $G$ from $C$: $d_G = L/2 - h = 0.300 - 0.260 = 0.040$ m (below $C$)
• Bullet point $B$ from $C$: $d_B = L - h = 0.600 - 0.260 = 0.340$ m

Moment of inertia of the bar about C

$$\bar I = \frac{1}{12}ML^2 = \frac{1}{12}(7)(0.6)^2 = 0.210\ \text{kg m}^2$$ $$I_C = \bar I + Md_G^2 = 0.210 + 7(0.040)^2 = 0.2212\ \text{kg m}^2$$

(a) Angular velocity just after impact

The bullet is embedded, so its speed after impact is $\omega d_B$. Conservation of angular momentum about $C$:

$$m_bv_0d_B = I_C\omega + m_b(\omega d_B)d_B$$ $$\omega = \frac{m_bv_0d_B}{I_C + m_bd_B^2} = \frac{0.04\times600\times0.340}{0.2212 + 0.04(0.340)^2} = 36.13\ \text{rad/s}$$

(b) Impulsive reaction at C

Take the bullet direction as positive. Linear impulse-momentum for the system (bar + bullet), the impulse of the reaction being $C\,\Delta t$:

$$m_bv_0 + C\,\Delta t = M(\omega d_G) + m_b(\omega d_B)$$ $$C\,\Delta t = 7(36.13)(0.040) + 0.04(36.13)(0.340) - 0.04(600) = -13.39\ \text{N s}$$ $$C = \frac{C\,\Delta t}{\Delta t} = \frac{-13.39}{0.001} = -13391\ \text{N}$$

The negative sign means the reaction at $C$ acts opposite to the bullet's direction.

Answer: (a) $\omega = 36.13$ rad/s; (b) $C \approx 13391$ N ($\approx 13.39$ kN) opposite to the bullet velocity.

Assumed geometry

Rod $AC$ is 0.600 m long: $A$ is 0.480 m above the pivot $B$, $C$ is 0.120 m below it. The fixed support $D$ is at the level of $B$, 0.350 m horizontally from $B$. The rod turns through $90^\circ$ so that it becomes horizontal and $C$ moves towards $D$.

            A
            |   0.48 m
            |
   D .......B (pivot)
     <0.35>  |   0.12 m
            C

Work-energy principle: $T_1 + V_1 = T_2 + V_2$ with $T_1 = 0$.

Spring

$$\begin{aligned} \text{Position 1: } CD_1 &= \sqrt{0.350^2 + 0.120^2} = 0.3700\ \text{m}, & x_1 &= 0.3700 - 0.150 = 0.2200\ \text{m}\\ \text{Position 2: } CD_2 &= 0.350 - 0.120 = 0.2300\ \text{m}, & x_2 &= 0.2300 - 0.150 = 0.0800\ \text{m} \end{aligned}$$ $$V_{e1} = \tfrac12(400)(0.2200)^2 = 9.68\ \text{J}, \qquad V_{e2} = \tfrac12(400)(0.0800)^2 = 1.28\ \text{J}$$

Gravity

The centre $G$ of the rod is $0.300$ m from $A$, i.e. $0.480 - 0.300 = 0.180$ m above $B$ initially, and $0.180$ m beside $B$ (level with it) at the end. So $G$ falls $0.180$ m:

$$W_g = mgh = 4(9.81)(0.180) = 7.06\ \text{J}$$

Moment of inertia about B

$$I_B = \frac{1}{12}(4)(0.6)^2 + 4(0.18)^2 = 0.2496\ \text{kg m}^2$$

Energy equation

$$0 + V_{e1} + V_{g1} = \tfrac12 I_B\omega^2 + V_{e2} + V_{g2}, \quad V_{g1}-V_{g2} = W_g \;\Rightarrow\; \tfrac12I_B\omega^2 = 9.68 - 1.28 + 7.06 = 15.46\ \text{J}$$ $$\omega = \sqrt{\frac{2(15.46)}{0.2496}} = 11.13\ \text{rad/s}$$

Answer: $\omega \approx 11.13$ rad/s (counter-clockwise, towards the spring support).

Impulsive motion is motion produced by very large forces acting for a very short time (like a hammer blow or a bullet strike), so that a finite change of velocity occurs in a negligible time. The product $\int F\,dt$ (the impulse) is finite; non-impulsive forces such as weights and spring forces are neglected during the impact, and the position of the body does not change.

Eccentric impact occurs when the line of impact (common normal at the point of contact) does not pass through the mass centre of one or both bodies, so the bodies rotate as well as translate after impact. The coefficient of restitution is applied to the relative velocities of the contact points along the line of impact, and the angular momentum principle is used about the pivot or mass centre.

Numerical part

Assumed geometry

Rod $AC$ is 0.600 m long: $A$ is 0.480 m above the pivot $B$, $C$ is 0.120 m below it. The fixed support $D$ is at the level of $B$, 0.350 m horizontally from $B$. The rod turns through $90^\circ$ so that it becomes horizontal and $C$ moves towards $D$.

            A
            |   0.48 m
            |
   D .......B (pivot)
     <0.35>  |   0.12 m
            C

Work-energy principle: $T_1 + V_1 = T_2 + V_2$ with $T_1 = 0$.

Spring

$$\begin{aligned} \text{Position 1: } CD_1 &= \sqrt{0.350^2 + 0.120^2} = 0.3700\ \text{m}, & x_1 &= 0.3700 - 0.150 = 0.2200\ \text{m}\\ \text{Position 2: } CD_2 &= 0.350 - 0.120 = 0.2300\ \text{m}, & x_2 &= 0.2300 - 0.150 = 0.0800\ \text{m} \end{aligned}$$ $$V_{e1} = \tfrac12(400)(0.2200)^2 = 9.68\ \text{J}, \qquad V_{e2} = \tfrac12(400)(0.0800)^2 = 1.28\ \text{J}$$

Gravity

The centre $G$ of the rod is $0.300$ m from $A$, i.e. $0.480 - 0.300 = 0.180$ m above $B$ initially, and $0.180$ m beside $B$ (level with it) at the end. So $G$ falls $0.180$ m:

$$W_g = mgh = 4(9.81)(0.180) = 7.06\ \text{J}$$

Moment of inertia about B

$$I_B = \frac{1}{12}(4)(0.6)^2 + 4(0.18)^2 = 0.2496\ \text{kg m}^2$$

Energy equation

$$0 + V_{e1} + V_{g1} = \tfrac12 I_B\omega^2 + V_{e2} + V_{g2}, \quad V_{g1}-V_{g2} = W_g \;\Rightarrow\; \tfrac12I_B\omega^2 = 9.68 - 1.28 + 7.06 = 15.46\ \text{J}$$ $$\omega = \sqrt{\frac{2(15.46)}{0.2496}} = 11.13\ \text{rad/s}$$

Answer: $\omega \approx 11.13$ rad/s (counter-clockwise, towards the spring support).

Impulse-momentum principle for 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

Data and choice of method

Energy method (work-energy): the rod starts from rest and the spring and gravity do work; the pivot reaction does none.

$L = 1.5$ m, $m = 15$ kg, $k = 300$ kN/m $= 3\times10^5$ N/m, compression $x = 0.025$ m. Pivot $O$ is 0.3 m from $B$, so the centre $G$ of the rod is

$$d = \frac{L}{2} - 0.3 = 0.75 - 0.3 = 0.45\ \text{m from } O \text{ on the } A \text{ side}$$

  B ===O=====G========== A
     0.3   0.45
                       ^ spring

Moment of inertia about $O$:

$$I_O = \frac{1}{12}mL^2 + md^2 = \frac{1}{12}(15)(1.5)^2 + 15(0.45)^2 = 5.850\ \text{kg m}^2$$

Position 1 (horizontal, at rest) and position 2 (vertical)

The spring is released, pushing end $A$ up, so the rod swings with $A$ up and $G$ rises by $d$ in reaching the vertical. $T_1 = 0$.

• Elastic energy released by the spring: $V_e = \tfrac12 kx^2 = \tfrac12(3\times10^5)(0.025)^2 = 93.75$ J
• Gain in potential energy of the rod: $V_g = mgd = 15(9.81)(0.45) = 66.22$ J

$$T_1 + V_1 = T_2 + V_2 \;\Rightarrow\; 0 + V_e = \tfrac12 I_O\omega^2 + V_g$$ $$\tfrac12 I_O\omega^2 = 93.75 - 66.22 = 27.53\ \text{J}$$ $$\omega = \sqrt{\frac{2(27.53)}{5.850}} = 3.07\ \text{rad/s}$$

Answer: $\omega \approx 3.07$ rad/s (counter-clockwise, with $A$ moving upward) as the rod passes the vertical.

Data and choice of method

Energy method (work-energy): the rod starts from rest and the spring and gravity do work; the pivot reaction does none.

$L = 1.5$ m, $m = 15$ kg, $k = 300$ kN/m $= 3\times10^5$ N/m, compression $x = 0.025$ m. Pivot $O$ is 0.3 m from $B$, so the centre $G$ of the rod is

$$d = \frac{L}{2} - 0.3 = 0.75 - 0.3 = 0.45\ \text{m from } O \text{ on the } A \text{ side}$$

  B ===O=====G========== A
     0.3   0.45
                       ^ spring

Moment of inertia about $O$:

$$I_O = \frac{1}{12}mL^2 + md^2 = \frac{1}{12}(15)(1.5)^2 + 15(0.45)^2 = 5.850\ \text{kg m}^2$$

Position 1 (horizontal, at rest) and position 2 (vertical)

The spring is released, pushing end $A$ up, so the rod swings with $A$ up and $G$ rises by $d$ in reaching the vertical. $T_1 = 0$.

• Elastic energy released by the spring: $V_e = \tfrac12 kx^2 = \tfrac12(3\times10^5)(0.025)^2 = 93.75$ J
• Gain in potential energy of the rod: $V_g = mgd = 15(9.81)(0.45) = 66.22$ J

$$T_1 + V_1 = T_2 + V_2 \;\Rightarrow\; 0 + V_e = \tfrac12 I_O\omega^2 + V_g$$ $$\tfrac12 I_O\omega^2 = 93.75 - 66.22 = 27.53\ \text{J}$$ $$\omega = \sqrt{\frac{2(27.53)}{5.850}} = 3.07\ \text{rad/s}$$

Answer: $\omega \approx 3.07$ rad/s (counter-clockwise, with $A$ moving upward) as the rod passes the vertical.

Data and assumptions

$L = 2.5$ m, $m=15$ kg, $k = 300$ kN/m, compression $x = 1$ in $= 0.0254$ m. Pivot $O$ is 0.5 m from end $B$ (taken as the end resting on the spring), so the centre $G$ is $d = 1.25 - 0.5 = 0.75$ m from $O$ on the free (long) side.

  spring               O                  G
  B ====================o================== A
   |<---- 0.5 m ---->|<------ 2.0 m ------>|

The spring pushes the short end up, so the long side swings down and $G$ falls by $d$ on reaching the vertical. (If the spring pressed the long end instead, the spring energy would be only 96.8 J, less than the 110.4 J needed to lift $G$ through 0.75 m, and the rod could not reach the vertical.)

$$I_O = \frac{1}{12}mL^2 + md^2 = \frac{1}{12}(15)(2.5)^2 + 15(0.75)^2 = 16.250\ \text{kg m}^2$$

Angular velocity (work-energy)

$$V_e = \tfrac12kx^2 = \tfrac12(3\times10^5)(0.0254)^2 = 96.77\ \text{J}, \qquad W_g = mgd = 15(9.81)(0.75) = 110.36\ \text{J}$$ $$T_1 + U_{1\to2} = T_2:\quad 0 + 96.77 + 110.36 = \tfrac12 I_O\omega^2$$ $$\omega = \sqrt{\frac{2(207.14)}{16.250}} = 5.05\ \text{rad/s}$$

Reaction at the pivot in the vertical position

At the vertical, $G$ is directly below $O$. The weight passes through $O$, so $\sum M_O = 0$ and $\alpha = 0$: there is no tangential acceleration and no horizontal reaction. $G$ has only the normal (centripetal) acceleration $\bar a_n = \omega^2 d$ directed up towards $O$:

$$\bar a_n = \omega^2 d = (5.05)^2(0.75) = 19.12\ \text{m/s}^2$$ $$\sum F_y = m\bar a_n:\quad R - mg = m\omega^2 d$$ $$R = 15(9.81 + 19.12) = 434\ \text{N}$$

Answer: $\omega \approx 5.05$ rad/s; reaction at $O$ is vertical, $R \approx 434$ N upward (horizontal component zero).

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 $v = \omega r$.

Impact problem

The rod hangs from the pin $A$ and the ball strikes its centre $G$, $d = 0.5$ m below $A$. During the impact the weights and the pin reaction (it passes through $A$) give no moment about $A$, so angular momentum about $A$ is conserved.

Data

$$m_r = \frac{50}{9.81} = 5.097\ \text{kg}, \qquad m_b = \frac{10}{9.81} = 1.019\ \text{kg}, \qquad I_A = \frac13 m_rL^2 = \frac13(5.097)(1)^2 = 1.699\ \text{kg m}^2$$

Let $v_b'$ be the ball's velocity after impact (positive in its original direction) and $\omega$ the angular velocity of the rod.

Conservation of angular momentum about A

$$m_b v\,d = I_A\omega + m_b v_b'\,d \quad (1)$$ $$1.019(9)(0.5) = 1.699\,\omega + 1.019\,v_b'(0.5)$$

Coefficient of restitution

Velocity of the rod at the point of impact is $\omega d$:

$$e = \frac{\omega d - v_b'}{v - 0} \;\Rightarrow\; 0.5\omega - v_b' = 0.4(9) = 3.6 \quad (2)$$

Solving

Substituting $v_b' = 0.5\omega - 3.6$ in (1):

$$4.587 = 1.699\omega + 0.5097(0.5\omega - 3.6) \;\Rightarrow\; \omega = \frac{4.587 + 0.5097(3.6)}{1.699 + 0.5097(0.5)} = 3.29\ \text{rad/s}$$

Then $v_b' = 0.5(3.29) - 3.6 = -1.96$ m/s.

Answer: angular velocity of the rod just after impact $\omega = 3.29$ rad/s; the ball rebounds with a velocity of $1.96$ m/s opposite to its original direction (the negative value $v_b' = -1.96$ m/s).

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

Impact problem

Take moments about the hinge $A$: during the impact the hinge reaction passes through $A$ and the weights are non-impulsive, so the angular momentum about $A$ of sphere + rod is conserved. Let $L = 1.2$ m be the distance from $A$ to the point of impact (the lower end $B$), $v_s'$ the sphere velocity after impact (positive in its original direction) and $\omega$ the angular velocity of the rod.

Data

$m_s = 3$ kg, $v_s = 5$ m/s, $m_r = 8$ kg, $e = 0.8$

$$I_A = \frac13 m_rL^2 = \frac13(8)(1.2)^2 = 3.840\ \text{kg m}^2$$

Conservation of angular momentum about A

$$m_sv_sL = I_A\omega + m_sv_s'L \quad (1)$$

Restitution

Velocity of the rod's lower end is $\omega L$:

$$e = \frac{\omega L - v_s'}{v_s} \;\Rightarrow\; v_s' = \omega L - e\,v_s \quad (2)$$

Solving

Substituting (2) in (1):

$$m_sv_sL(1+e) = (I_A + m_sL^2)\,\omega$$ $$\omega = \frac{3(5)(1.2)(1+0.8)}{3.840 + 3(1.2)^2} = 3.97\ \text{rad/s}$$ $$v_s' = \omega L - e\,v_s = 3.97(1.2) - 0.8(5) = 0.76\ \text{m/s}$$

Answer: $\omega = 3.97$ rad/s; the sphere's velocity just after impact is $0.76$ m/s (in the original direction).

Take moments about the hinge $A$: during the impact the hinge reaction passes through $A$ and the weights are non-impulsive, so the angular momentum about $A$ of sphere + rod is conserved. Let $L = 1.2$ m be the distance from $A$ to the point of impact (the lower end $B$), $v_s'$ the sphere velocity after impact (positive in its original direction) and $\omega$ the angular velocity of the rod.

(The figure label shows 5 m/s but the text gives 7 m/s; 7 m/s is used.)

Data

$m_s = 2.5$ kg, $v_s = 7$ m/s, $m_r = 10$ kg, $e = 0.89$

$$I_A = \frac13 m_rL^2 = \frac13(10)(1.2)^2 = 4.800\ \text{kg m}^2$$

Conservation of angular momentum about A

$$m_sv_sL = I_A\omega + m_sv_s'L \quad (1)$$

Restitution

Velocity of the rod's lower end is $\omega L$:

$$e = \frac{\omega L - v_s'}{v_s} \;\Rightarrow\; v_s' = \omega L - e\,v_s \quad (2)$$

Solving

Substituting (2) in (1):

$$m_sv_sL(1+e) = (I_A + m_sL^2)\,\omega$$ $$\omega = \frac{2.5(7)(1.2)(1+0.89)}{4.800 + 2.5(1.2)^2} = 4.73\ \text{rad/s}$$ $$v_s' = \omega L - e\,v_s = 4.73(1.2) - 0.89(7) = -0.56\ \text{m/s}$$

Answer: $\omega = 4.73$ rad/s; the sphere's velocity just after impact is $-0.56$ m/s (reversed).

Method. Work-energy for the whole system: the couple does work $M\theta_C$; friction is neglected, and the weights and bearing reactions do no work. Gears $A$ and $B$ are identical and are driven by $C$, so $\omega_A = \omega_B = \omega_C\,\dfrac{r_C}{r_A}$.

Data

$m_A = m_B = 2.1$ kg, $\bar k_A = 0.103$ m, $m_C = 10.3$ kg, $\bar k_C = 0.187$ m, $r_A = r_B = 0.107$ m, $r_C = 0.265$ m, $M = 9.26$ N m

$$\bar I_A = \bar I_B = 2.1(0.103)^2 = 0.0223\ \text{kg m}^2, \qquad \bar I_C = 10.3(0.187)^2 = 0.3602\ \text{kg m}^2$$ $$\frac{r_C}{r_A} = \frac{0.265}{0.107} = 2.4766$$

        (B)
          \
    M -> ( C ) 
          /
        (A)

Angular speeds

$$\omega_1 = 100\ \text{rpm} = 10.47\ \text{rad/s}, \qquad \omega_2 = 450\ \text{rpm} = 47.12\ \text{rad/s}$$

(i) Revolutions of gear C

Kinetic energy in terms of $\omega_C$:

$$T = \tfrac12\bar I_C\omega_C^2 + 2\left(\tfrac12\bar I_A\omega_A^2\right) = \tfrac12\left[\bar I_C + 2\bar I_A\left(\frac{r_C}{r_A}\right)^2\right]\omega_C^2 = \tfrac12 I_{eq}\,\omega_C^2$$ $$I_{eq} = 0.3602 + 2(0.0223)(2.4766)^2 = 0.6335\ \text{kg m}^2$$ $$T_1 + M\theta_C = T_2 \;\Rightarrow\; \theta_C = \frac{I_{eq}(\omega_2^2 - \omega_1^2)}{2M} = \frac{0.6335\,(47.12^2 - 10.47^2)}{2(9.26)} = 72.21\ \text{rad}$$ $$\text{revolutions} = \frac{\theta_C}{2\pi} = \frac{72.21}{2\pi} = 11.49$$

(ii) Tangential force on gear A

The couple is constant, so the angular acceleration is constant:

$$\alpha_C = \frac{M}{I_{eq}} = \frac{9.26}{0.6335} = 14.62\ \text{rad/s}^2, \qquad \alpha_A = \alpha_C\frac{r_C}{r_A} = 36.20\ \text{rad/s}^2$$

For gear $A$ the only moment about its axis is that of the tangential tooth force $F$ (friction neglected):

$$F\,r_A = \bar I_A\alpha_A \;\Rightarrow\; F = \frac{0.0223\times36.20}{0.107} = 7.54\ \text{N}$$

(Check on gear $C$: $M - 2Fr_C = \bar I_C\alpha_C$ gives $9.26 - 2(7.54)(0.265) = 5.26$ and $\bar I_C\alpha_C = 5.26$.)

Answer: (i) $11.49$ revolutions of gear $C$; (ii) $F = 7.54$ N.

(Weights are given in kg, so they are used as masses.)

Method. Work-energy for the whole system: the couple does work $M\theta_C$; friction is neglected, and the weights and bearing reactions do no work. Gears $A$ and $B$ are identical and are driven by $C$, so $\omega_A = \omega_B = \omega_C\,\dfrac{r_C}{r_A}$.

Data

$m_A = m_B = 2.5$ kg, $\bar k_A = 0.1$ m, $m_C = 12.5$ kg, $\bar k_C = 0.18$ m, $r_A = r_B = 0.1$ m, $r_C = 0.25$ m, $M = 10$ N m

$$\bar I_A = \bar I_B = 2.5(0.1)^2 = 0.0250\ \text{kg m}^2, \qquad \bar I_C = 12.5(0.18)^2 = 0.4050\ \text{kg m}^2$$ $$\frac{r_C}{r_A} = \frac{0.25}{0.1} = 2.5000$$

        (B)
          \
    M -> ( C ) 
          /
        (A)

Angular speeds

$$\omega_1 = 100\ \text{rpm} = 10.47\ \text{rad/s}, \qquad \omega_2 = 450\ \text{rpm} = 47.12\ \text{rad/s}$$

(i) Revolutions of gear C

Kinetic energy in terms of $\omega_C$:

$$T = \tfrac12\bar I_C\omega_C^2 + 2\left(\tfrac12\bar I_A\omega_A^2\right) = \tfrac12\left[\bar I_C + 2\bar I_A\left(\frac{r_C}{r_A}\right)^2\right]\omega_C^2 = \tfrac12 I_{eq}\,\omega_C^2$$ $$I_{eq} = 0.4050 + 2(0.0250)(2.5000)^2 = 0.7175\ \text{kg m}^2$$ $$T_1 + M\theta_C = T_2 \;\Rightarrow\; \theta_C = \frac{I_{eq}(\omega_2^2 - \omega_1^2)}{2M} = \frac{0.7175\,(47.12^2 - 10.47^2)}{2(10)} = 75.73\ \text{rad}$$ $$\text{revolutions} = \frac{\theta_C}{2\pi} = \frac{75.73}{2\pi} = 12.05$$

(ii) Tangential force on gear A

The couple is constant, so the angular acceleration is constant:

$$\alpha_C = \frac{M}{I_{eq}} = \frac{10}{0.7175} = 13.94\ \text{rad/s}^2, \qquad \alpha_A = \alpha_C\frac{r_C}{r_A} = 34.84\ \text{rad/s}^2$$

For gear $A$ the only moment about its axis is that of the tangential tooth force $F$ (friction neglected):

$$F\,r_A = \bar I_A\alpha_A \;\Rightarrow\; F = \frac{0.0250\times34.84}{0.1} = 8.71\ \text{N}$$

(Check on gear $C$: $M - 2Fr_C = \bar I_C\alpha_C$ gives $10 - 2(8.71)(0.25) = 5.64$ and $\bar I_C\alpha_C = 5.64$.)

Answer: (i) $12.05$ revolutions of gear $C$; (ii) $F = 8.71$ N.

Data. $r_A=0.250$ m, $m_A = 10$ kg, $\bar k_A = 0.200$ m; $r_B = 0.100$ m, $m_B=3$ kg, $\bar k_B = 0.080$ m; $M = 8$ N m on $B$; the system starts from rest.

$$\bar I_A = 10(0.200)^2 = 0.400\ \text{kg m}^2, \qquad \bar I_B = 3(0.080)^2 = 0.0192\ \text{kg m}^2$$

The teeth do not slip, so $\omega_A r_A = \omega_B r_B$:

$$\omega_A = \omega_B\frac{r_B}{r_A} = 0.4\,\omega_B$$

(a) Revolutions of gear B (work-energy)

$$T = \tfrac12\bar I_B\omega_B^2 + \tfrac12\bar I_A\omega_A^2 = \tfrac12\left[\bar I_B + \bar I_A\left(\frac{r_B}{r_A}\right)^2\right]\omega_B^2 = \tfrac12(0.0832)\,\omega_B^2$$ $$\bar I_B + \bar I_A(0.4)^2 = 0.0192 + 0.400(0.16) = 0.0832\ \text{kg m}^2$$ $$\omega_B = 540\ \text{rpm} = 56.55\ \text{rad/s}$$ $$T_1 + M\theta_B = T_2:\quad \theta_B = \frac{\tfrac12(0.0832)(56.55)^2}{8} = 16.63\ \text{rad} = 2.65\ \text{rev}$$

(b) Tangential force of B on A

Angular acceleration (constant couple):

$$\alpha_B = \frac{M}{I_{eq}} = \frac{8}{0.0832} = 96.15\ \text{rad/s}^2$$

For gear $B$: $M - F r_B = \bar I_B\alpha_B$

$$F = \frac{M - \bar I_B\alpha_B}{r_B} = \frac{8 - 0.0192(96.15)}{0.100} = 61.54\ \text{N}$$

(Check with gear $A$: $\alpha_A = 0.4\alpha_B = 38.46$ rad/s², $F = \bar I_A\alpha_A/r_A = 61.54$ N.)

Answer: (a) $\theta_B = 16.63$ rad $= 2.65$ revolutions; (b) $F = 61.54$ N.

Solid sphere of weight $W = 80$ kN, rolling without slipping on a $30^\circ$ incline: $\bar I = \tfrac25 mr^2$ and $\bar a = \alpha r$. Distances are in feet, so $g = 32.2$ ft/s².

(a) Minimum coefficient of friction

Take axes along and perpendicular to the incline. Let $F$ be friction up the incline.

$$\begin{aligned} W\sin30^\circ - F &= m\bar a\\ N - W\cos30^\circ &= 0\\ Fr &= \bar I\alpha = \tfrac25 mr^2\frac{\bar a}{r} \;\Rightarrow\; F = \tfrac25 m\bar a \end{aligned}$$

Substituting $F$ in the first equation: $W\sin30^\circ = \tfrac75 m\bar a$, so

$$\bar a = \tfrac57 g\sin30^\circ = \tfrac57(32.2)(0.5) = 11.50\ \text{ft/s}^2$$ $$F = \tfrac25 m\bar a = \tfrac27W\sin30^\circ = \tfrac27(80)(0.5) = 11.43\ \text{kN}, \qquad N = W\cos30^\circ = 80\cos30^\circ = 69.28\ \text{kN}$$

For rolling without slipping, $F \le \mu_s N$:

$$\mu_{min} = \frac{F}{N} = \tfrac27\tan30^\circ = 0.1650$$

(b) Velocity of G after rolling 15 ft

The acceleration is constant, starting from rest:

$$\bar v^2 = 2\bar a\,s = 2(11.50)(15) \;\Rightarrow\; \bar v = 18.57\ \text{ft/s} \;(= 5.66\ \text{m/s})$$

(Check by energy: $Ws\sin30^\circ = \tfrac12m\bar v^2 + \tfrac12\bar I\omega^2 = \tfrac12\cdot\tfrac75 m\bar v^2$, giving the same $\bar v$.)

(c) Frictionless incline, 20 ft

The sphere does not rotate; it slides with $a = g\sin30^\circ = 16.1$ ft/s²:

$$\bar v = \sqrt{2(16.1)(20)} = 25.38\ \text{ft/s} \;(= 7.73\ \text{m/s})$$

Answer: (a) $\mu_{min} = 0.1650$; (b) $\bar v = 18.57$ ft/s; (c) $\bar v = 25.38$ ft/s.

Central and eccentric impact

Motion of the two rods

Rods $AB$ and $BD$ are identical ($L=0.75$ m, $m=6$ kg), joined by a pin at $B$. $A$ is a fixed pin and $D$ slides on a smooth horizontal surface (rollers). Released from rest at $\beta=50^\circ$, the system moves under gravity only, so use the work-energy principle.

          B
         / \
        /   \
       /     \
    A /_______\ D   (rollers)
       beta

Kinematics

Let $\omega_{AB}$ be the angular velocity of $AB$. $\vec v_B \perp AB$ and $\vec v_D$ is horizontal. The ICR $I$ of $BD$ lies on the line $AB$ extended and on the vertical through $D$. By symmetry of the triangle $ABD$, $IB = BD = L$ and $ID = 2L\sin\beta$, so

$$\omega_{BD} = \frac{v_B}{IB} = \frac{\omega_{AB}L}{L} = \omega_{AB} = \omega$$

(the two rods turn in opposite senses). The centre $G_2$ of $BD$ is at $(1.5L\cos\beta,\ 0.5L\sin\beta)$ and $I$ is at $(2L\cos\beta,\ 2L\sin\beta)$, so

$$\left(\frac{IG_2}{L}\right)^2 = 0.25\cos^2\beta + 2.25\sin^2\beta$$

At $\beta = 20^\circ$: $0.25\cos^2 20^\circ + 2.25\sin^2 20^\circ = 0.4840$, so $IG_2 = 0.5218$ m.

Kinetic energy at $\beta = 20^\circ$

$$T_{AB} = \tfrac12\left(\tfrac13mL^2\right)\omega^2 = \tfrac12(1.1250)\,\omega^2$$ $$T_{BD} = \tfrac12\left[\tfrac1{12}mL^2 + m(IG_2)^2\right]\omega^2 = \tfrac12(1.9146)\,\omega^2$$ $$T_2 = \tfrac12(1.1250 + 1.9146)\,\omega^2 = 1.5198\,\omega^2$$

Work of gravity

Heights of the mass centres: $\tfrac L2\sin\beta$ for each rod, so the total potential energy is $V = mgL\sin\beta$.

$$U_{1\to2} = mgL(\sin50^\circ - \sin20^\circ) = 6(9.81)(0.75)(0.7660 - 0.3420) = 18.72\ \text{J}$$

(a) Angular velocity of AB

$$T_1 + U_{1\to2} = T_2:\quad 0 + 18.72 = 1.5198\,\omega^2 \;\Rightarrow\; \omega_{AB} = 3.51\ \text{rad/s}$$

AB turns clockwise (B moving down and to the right).

(b) Velocity of D

$D$ is $ID = 2L\sin20^\circ$ from the ICR:

$$v_D = \omega_{BD}\,(ID) = 3.51\times2(0.75)\sin20^\circ = 3.51\times0.5130 = 1.80\ \text{m/s}\ (\to)$$

Answer: $\omega_{AB} = 3.51$ rad/s clockwise; $v_D = 1.80$ m/s horizontally to the right.