Please put the on the epicyclic gears train table

When the arm is fixed, consider the relative speeds:

• Gear B (internal) drives Gear C (external). They rotate in the same direction. $\frac{\omega_C}{\omega_B} = \frac{N_B}{N_C} = \frac{92}{25}$
• Gear D (external) drives Gear E (external). They rotate in opposite directions. $\frac{\omega_E}{\omega_D} = -\frac{N_D}{N_E} = -\frac{15}{52}$
• Since C and D are compounded, $\omega_C = \omega_D$. So, $\frac{\omega_E}{\omega_B} = \frac{\omega_C}{\omega_B} \times \frac{\omega_E}{\omega_D} = \frac{N_B}{N_C} \times \left(-\frac{N_D}{N_E}\right) = \frac{92}{25} \times \left(-\frac{15}{52}\right)$ $\frac{\omega_E}{\omega_B} = -\frac{92 \times 15}{25 \times 52} = -\frac{1380}{1300} = -\frac{69}{65}$

Let $x$ be the revolutions of gear B when the arm is fixed, and $y$ be the revolutions of the arm.

Condition	Arm A	Gear B	Gear C	Gear D	Gear E
1. Arm fixed, Gear B rotates by $+x$	0	$+x$	$+x \frac{N_B}{N_C}$	$+x \frac{N_B}{N_C}$	$-x \frac{N_B}{N_C} \frac{N_D}{N_E}$
			$+x \frac{92}{25}$	$+x \frac{92}{25}$	$-x \frac{69}{65}$
2. Add $+y$ revolutions to all components	$+y$	$+y$	$+y$	$+y$	$+y$
3. Total motion (Sum of 1 and 2)	$y$	$x+y$	$x \frac{92}{25} + y$	$x \frac{92}{25} + y$	$-x \frac{69}{65} + y$

From the problem:

• Speed of Arm A ($\omega_A$) = 130 rpm. So, $y = 130$.
• Speed of Gear B ($\omega_B$) = 0 rpm (fixed). So, $x+y = 0$.

Substitute $y=130$ into $x+y=0$:

$$x + 130 = 0$$ $$x = -130$$

Using the total motion for Gear E:

$$\omega_E = -x \frac{69}{65} + y$$

Substitute $x = -130$ and $y = 130$:

$$\omega_E = -(-130) \frac{69}{65} + 130$$ $$\omega_E = 130 \times \frac{69}{65} + 130$$ $$\omega_E = (2 \times 65) \times \frac{69}{65} + 130$$ $$\omega_E = 2 \times 69 + 130$$ $$\omega_E = 138 + 130$$ $$\omega_E = 268 \text{ rpm}$$

The speed of wheel E is $\text{268 rpm}$.