Inertia force, inertia torque and D'Alembert's principle

(a) Definitions

• Centrifugal force: the outward radial force on a body in circular motion, $F_c = m\omega^2 r$. It is the reaction to the centripetal force. For analysis it is taken as an inertia force.
• Inertia force: the fictitious force equal to $m a_G$ acting at the centre of mass G, opposite to the acceleration of G: $\vec F_I = -m\vec a_G$.
• Inertia torque (couple): equal to $I_G\alpha$, opposite to the angular acceleration $\alpha$: $C_I = -I_G\alpha$.
• D'Alembert's principle: if the inertia force and inertia torque are added to the actual external forces on a body, the body is in dynamic equilibrium. Then $\sum \vec F + \vec F_I = 0$ and $\sum M + C_I = 0$, so the methods of statics can be used for dynamic problems.

Numerical.

$$\omega = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s}$$ $$F_c = m\omega^2r = 5 \times 986.96 \times 0.4 = 1973.9\ \text{N}$$

(b) Link

Inertia force at G:

$$F_I = m a_G = 12 \times 25 = 300\ \text{N}$$

directed opposite to the acceleration of G.

Inertia couple:

$$C_I = I_G\alpha = 0.8 \times 40 = 32\ \text{N m}$$

acting clockwise (opposite to $\alpha$).

Single equivalent force. The force $F_I$ and couple $C_I$ can be replaced by the same force $F_I$ acting along a line parallel to its original line, shifted by

$$h = \frac{C_I}{F_I} = \frac{32}{300} = 0.1067\ \text{m} = 106.7\ \text{mm}$$

from G, on the side that gives a moment about G in the same sense as $C_I$ (clockwise).

         h
    ----------|-> F_I (shifted line)
   G (x)  <- C_I clockwise

Answer: (a) $F_c = 1973.9$ N; (b) $F_I = 300$ N, $C_I = 32$ N m (clockwise), shifted force at $h = 106.7$ mm from G.