Gears A, B, C with couple, revolutions and force

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)
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    M -> ( C ) 
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        (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.