Plane Kinetics of Rigid Bodies: Force, Mass and Acceleration

Definitions

Mass moment of inertia of a body about an axis is $I = \int r^2\,dm$, where $r$ is the perpendicular distance of the element $dm$ from the axis (unit kg·m²). It measures resistance to angular acceleration, as mass does for linear acceleration.

Radius of gyration $k$ is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia: $I = mk^2$, so $k = \sqrt{I/m}$.

Parallel axis theorem

Statement: the moment of inertia about any axis equals the moment of inertia about a parallel axis through the mass centre plus $md^2$:

$$I = I_G + md^2$$

Proof: take the mass centre G as origin of x-y axes, with the axis G along z and the parallel axis passing through the point $(d, 0)$. For an element $dm$ at $(x,y)$:

$$I = \int [(x - d)^2 + y^2]\,dm = \int (x^2 + y^2)\,dm - 2d\int x\,dm + d^2\int dm$$

The first term is $I_G$. The second is zero because $\int x\,dm = m\bar x = 0$ (G is the origin). The third is $md^2$. Hence $I = I_G + md^2$.

(a) Slender rod, length L, mass m

Mass per unit length $= m/L$. About the transverse axis through the centre:

$$I_G = \int_{-L/2}^{L/2} x^2 \frac{m}{L}\,dx = \frac{m}{L}\cdot\frac{2}{3}\left(\frac{L}{2}\right)^3 = \frac{mL^2}{12}$$

About one end (parallel axis theorem, $d = L/2$):

$$I_{end} = \frac{mL^2}{12} + m\left(\frac{L}{2}\right)^2 = \frac{mL^2}{3}$$

(b) Solid cylinder, radius R, length h, density $\rho$

Take a thin cylindrical shell of radius $r$ and thickness $dr$: $dm = \rho\,(2\pi r\,h)\,dr$.

$$I = \int_0^R r^2\,\rho\,2\pi h\,r\,dr = 2\pi\rho h\,\frac{R^4}{4} = \frac{1}{2}(\rho\pi R^2 h)R^2 = \frac{1}{2}mR^2$$

(independent of the length h).

Given: rod $m_r = 3$ kg, $L = 1.2$ m; sphere $m_s = 2$ kg, $R = 0.1$ m, centre on the rod axis at $1.2 + 0.1 = 1.3$ m from the free end O.

 O                               ___
 |============================( o )
 |<------- 1.2 m ------------>|<R>
 |<------ 1.3 m to sphere centre

Moment of inertia about O (transverse axis)

Rod about its end: $I_r = \dfrac{m_rL^2}{3} = \dfrac{3(1.2)^2}{3} = 1.440$ kg·m²

Sphere about its own centre: $\tfrac25 m_sR^2 = 0.4 \times 2 \times 0.01 = 0.008$ kg·m²

Parallel axis shift: $m_s d^2 = 2(1.3)^2 = 3.380$ kg·m²

Centre of mass from O

$$\bar x = \frac{3(0.6) + 2(1.3)}{5} = \frac{1.8 + 2.6}{5} = 0.88\ \text{m}$$

Radius of gyration about O

$$k_O = \sqrt{\frac{I_O}{m}} = \sqrt{\frac{4.828}{5}} = 0.983\ \text{m}$$

(As a check, the moment of inertia about the centre of mass is $I_G = 4.828 - 5(0.88)^2 = 0.956$ kg·m².)

Answer: I_O = 4.828 kg·m²; centre of mass 0.88 m from the free end; k_O = 0.983 m.

For a rigid body of mass $m$ in plane motion with mass centre G, acceleration $\mathbf{a}_G$ and angular acceleration $\alpha$ (D'Alembert-Newton-Euler form), under external forces and couples in the plane:

$$\Sigma F_x = m\,a_{Gx} \qquad \Sigma F_y = m\,a_{Gy} \qquad \Sigma M_G = I_G\,\alpha$$

The first two give the motion of the mass centre (it moves as if all the mass and forces were at G); the third gives the rotation about G. $I_G$ is the mass moment of inertia about the axis through G perpendicular to the plane. As an alternative, the inertia force $-m\mathbf{a}_G$ (through G) and the inertia couple $-I_G\alpha$ can be added and the body treated in static equilibrium.

(a) Translation

$\alpha = 0$, so $\Sigma M_G = 0$ and the resultant force passes through G: $\Sigma F = m\mathbf{a}_G$. Example: a crate sliding on a surface (check tipping with $\Sigma M_G = 0$).

(b) Rotation about a fixed axis O

G moves in a circle, with $a_{Gt} = \alpha\bar r$ and $a_{Gn} = \omega^2\bar r$. Taking moments about O:

$$\Sigma M_O = I_O\,\alpha \qquad I_O = I_G + m\bar r^2$$

The reactions at O follow from $\Sigma F_t = m\bar r\alpha$ and $\Sigma F_n = m\bar r\omega^2$.

(c) General plane motion

Use the three equations with $a_G$ and $\alpha$ unknown, plus kinematic relations (for example, rolling without slipping $a_G = \alpha r$). Moments may be taken about any point P if the moment of $m\mathbf{a}_G$ is included:

$$\Sigma M_P = I_G\alpha + (\text{moment of } m\mathbf{a}_G \text{ about P})$$

For a body rolling without slipping on a fixed surface, taking P as the contact point C gives $\Sigma M_C = I_C\alpha$ only when G moves parallel to the surface (such as a homogeneous round body on a plane).

Procedure

1. Draw the free-body diagram with all forces and the unknown reactions.
2. Show $m\mathbf{a}_G$ and $I_G\alpha$ in a kinetic diagram.
3. Write the three equations and add kinematic constraint equations.
4. Solve for the unknowns.

Given: $m = 20$ kg, $r = 0.3$ m, $\theta = 30^\circ$, rolling without slipping. $I_G = \tfrac12 mr^2 = 0.90$ kg·m².

         \
          \  O (G)
           (o)--> a
            \ F (friction up slope)
             \
              \ 30 deg

Equations of motion

Let $F$ be the friction force up the slope at the contact point, $N$ the normal reaction, and $a$ the acceleration of G down the slope.

Along the slope: $mg\sin\theta - F = ma$ (1)

Perpendicular to the slope: $N = mg\cos\theta$ (2)

Moments about G: $F r = I_G\alpha$ (3)

Rolling without slipping: $a = \alpha r$ (4)

From (3) and (4): $F = \dfrac{I_G a}{r^2} = \tfrac12 ma$. Put in (1):

$$mg\sin\theta = ma + \tfrac12 ma = \tfrac32 ma \;\Rightarrow\; a = \tfrac23 g\sin\theta$$

(a) Linear acceleration

$$a = \tfrac23(9.81)(0.5) = 3.27\ \text{m/s}^2$$

(b) Angular acceleration

$$\alpha = \frac{a}{r} = \frac{3.27}{0.3} = 10.90\ \text{rad/s}^2$$

(c) Friction and normal reaction

$$F = \tfrac12 ma = \tfrac12(20)(3.27) = 32.7\ \text{N} \qquad N = 20(9.81)\cos30^\circ = 169.9\ \text{N}$$

(d) Minimum coefficient of friction

For no slipping $F \le \mu_s N$:

$$\mu_{min} = \frac{F}{N} = \frac{32.7}{169.9} = 0.1925 \;\left(= \tfrac13\tan30^\circ\right)$$

Answer: a = 3.27 m/s²; α = 10.90 rad/s²; F = 32.7 N; N = 169.9 N; μ_min = 0.192.

Given: $m_A = 8$ kg, $m_B = 4$ kg, pulley $M = 10$ kg, $r = 0.2$ m. Since $m_A > m_B$, A moves down and B moves up. Pulley moment of inertia:

$$I = \tfrac12 Mr^2 = \tfrac12(10)(0.2)^2 = 0.20\ \text{kg·m}^2$$

        ____
       /    \   pulley (M, r)
  T_A |      | T_B
      |      |
     [A]    [B]
   (down)   (up)

The rope does not slip, so $a = \alpha r$. The tensions differ in the two parts because the pulley has inertia.

Equations

Block A (downward positive): $m_Ag - T_A = m_A a$ (1)

Block B (upward positive): $T_B - m_Bg = m_B a$ (2)

Pulley (moments about axle): $(T_A - T_B)\,r = I\alpha = \tfrac12 Mr^2\,\dfrac{a}{r}$

$$T_A - T_B = \tfrac12 Ma = 5a \quad (3)$$

Adding (1), (2) and (3): $(m_A - m_B)g = (m_A + m_B + \tfrac12 M)\,a$

Acceleration

$$a = \frac{(8 - 4)(9.81)}{8 + 4 + 5} = \frac{39.24}{17} = 2.308\ \text{m/s}^2$$

Angular acceleration

$$\alpha = \frac{a}{r} = \frac{2.308}{0.2} = 11.54\ \text{rad/s}^2\ \text{(clockwise, toward A's side)}$$

Tensions

$$T_A = m_A(g - a) = 8(9.81 - 2.308) = 60.01\ \text{N}$$ $$T_B = m_B(g + a) = 4(9.81 + 2.308) = 48.47\ \text{N}$$

Check: $T_A - T_B = 11.54 = 5a = 11.54$ N. It agrees.

Answer: a = 2.31 m/s²; α = 11.54 rad/s²; T_A = 60.0 N; T_B = 48.5 N.

Given: $m = 6$ kg, $L = 1.5$ m, hinge at A, released from rest in the horizontal position.

   A (hinge) o==========o B
             |-- G --|
   W = mg  v  (at L/2)

At the instant of release the rod is at rest, so $\omega = 0$ and the centripetal term $\omega^2 (L/2)$ vanishes. The mass centre G has only a tangential (vertical) acceleration.

Moment of inertia about A

$$I_A = \frac{mL^2}{3} = \frac{6(1.5)^2}{3} = 4.50\ \text{kg·m}^2$$

Angular acceleration

Moments about the hinge (only the weight has a moment; the hinge reaction passes through A):

$$mg\,\frac{L}{2} = I_A\alpha \;\Rightarrow\; \alpha = \frac{44.14}{4.50} = 9.81\ \text{rad/s}^2\ \left(=\frac{3g}{2L}\right)$$

Acceleration of G

$$a_G = \alpha\,\frac{L}{2} = 9.81 \times 0.75 = 7.357\ \text{m/s}^2\ \text{(downward)}$$

Hinge reaction

Horizontal: $A_x = 0$ (no horizontal acceleration or force).

Vertical (downward positive): $mg - A_y = m\,a_G$

$$A_y = 6(9.81 - 7.357) = 14.72\ \text{N (upward)}$$

This is one quarter of the weight ($mg/4 = 14.71$ N), a known result for a uniform rod released from the horizontal position.

Answer: α = 9.81 rad/s²; a_G = 7.36 m/s² downward; hinge reaction = 14.7 N upward (A_x = 0).