Gears A and B, M = 8 N-m, revolutions

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.