- #26

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Free body diagram of pinion (A):A constant torque is applied to a pinion which has a moment of inertia of I_m. The pinion(A) drives two gears, one (B) which is connected to a mass which has a moment of inertia = I_m and the other(C) is connected to a mass which has a moment of inertia = 2I_m. The gear ration R_1=A/B is fixed and is equal to 3. What should the gear ration R_2=A/C be to give the maximum angular acceleration of gear 4? Neglect the mass of the gears.

- A constant torque ##T## is applied;
- A force ##F_B## is reacted at radius ##A## from the gear (B);
- A force ##F_C## is reacted at radius ##A## from the gear (C);
- Pinion (A) has moment of inertia ##I_m## and an angular acceleration ##\alpha_A##.

$$T = F_B A + F_C A + I_m\alpha_A$$

You have 3 unknowns (##F_B##, ##F_C## & ##\alpha_A##) therefore you need 3 equations to solve the problem.

Free body diagram of gear (B):

- A force ##F_B## is applied at radius ##B## from the pinion (A);
- Gear (B) has moment of inertia ##I_m## and an angular acceleration ##\alpha_B##.

$$F_B B = I_m\alpha_B$$

We introduce a 4th unknown (##\alpha_B##) thus we will need 4 equations.

Free body diagram of gear (C):

- A force ##F_C## is applied at radius ##C## from the pinion (A);
- Gear (C) has moment of inertia ##2I_m## and an angular acceleration ##\alpha_C##.

$$F_C C = 2I_m\alpha_C$$

We introduce a 5th unknown (##\alpha_C##) thus we will need 5 equations.

We also know that linear accelerations are the same for each gear at the contact point between pinion (A) and gear (B), thus:

$$a_A = a_B$$

$$\alpha_A A = \alpha_B B$$

Similarly for the linear accelerations between pinion (A) and gear (C):

$$\alpha_A A = \alpha_C C$$

There, we have 5 equations with 5 unknowns. So:

$$T = F_B A + F_C A + I_m\alpha_A$$

$$T = \left(\frac{I_m}{B} \alpha_B\right) A + \left(\frac{2I_m}{C} \alpha_C \right) A + I_m\alpha_A$$

$$T = \frac{A}{B} I_m\alpha_B + \frac{A}{C} 2I_m \alpha_C + I_m\alpha_A$$

$$T = \frac{A}{B} I_m\left(\frac{A}{B}\alpha_A\right) + \frac{A}{C} 2I_m \left(\frac{A}{C}\alpha_A\right) + I_m\alpha_A$$

$$T = \left(\frac{A}{B}\right)^2 I_m \alpha_A +\left(\frac{A}{C}\right)^2 2I_m \alpha_A + I_m\alpha_A$$

$$\alpha_A = \frac{1}{\left(\frac{A}{B}\right)^2 + 2\left(\frac{A}{C}\right)^2 + 1 }\frac{T}{I_m}$$

And then from the last 2 equations:

$$\alpha_B = \frac{A}{B}\alpha_A = \frac{^A/_B}{\left(\frac{A}{B}\right)^2 + 2\left(\frac{A}{C}\right)^2 + 1 }\frac{T}{I_m}$$

$$\alpha_C = \frac{A}{C}\alpha_A = \frac{^A/_C}{\left(\frac{A}{B}\right)^2 + 2\left(\frac{A}{C}\right)^2 + 1 }\frac{T}{I_m}$$

Now we can find where ##\alpha_C## will be max when its derivative will be zero, i.e.:

$$\frac{d\alpha_C}{d^A/_C} = \frac{\frac{\left(^A/_B\right)^2 + 1}{\left(^A/_C\right)^2} - 2}{\left( \frac{\left(^A/_B\right)^2 + 1}{^A/_C} + 2\ ^A/_C \right)^2}\frac{T}{I_m} = 0$$

Which means that:

$$^A/_C = \sqrt{\frac{\left(^A/_B\right)^2 + 1}{2}}$$

Or ##^A/_C = \sqrt{5} = 2.236##. Now, I know you said the answer was supposed ##\sqrt{1.8}##, but I'm pretty sure of my answer with the problem as stated (although I cannot be sure of what you meant by «gear 4»).