Two wheels with fixed hubs, each having a mass of $1 \text{ kg}$, start from rest, and forces are applied as shown.  Assume the hubs and spokes are massless, so that the rotational inertia is $I = mR^2$.  In order to impart identical angular accelerations about their respective hubs, how large must $F_2$ be?

[asy]
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
draw(circle((0,0),0.5));
draw((1, 0.5)--(0,0.5)--(0,-0.5),BeginArrow);
draw((-0.5,0)--(0.5,0));
draw((-0.5*sqrt(2)/2, 0.5*sqrt(2)/2)--(0.5*sqrt(2)/2,-0.5*sqrt(2)/2));
draw((0.5*sqrt(2)/2, 0.5*sqrt(2)/2)--(-0.5*sqrt(2)/2,-0.5*sqrt(2)/2));
label("$R$ = 0.5 m", (0, -0.5),S);
label("$F_1$ = 1 N",(1,0.5),N);
draw(circle((3,0.5),1));
draw((4.5,1.5)--(3,1.5)--(3,-0.5),BeginArrow);
draw((2,0.5)--(4,0.5));
draw((3-sqrt(2)/2, 0.5+sqrt(2)/2)--(3+sqrt(2)/2, 0.5-sqrt(2)/2));
draw((3+sqrt(2)/2, 0.5+sqrt(2)/2)--(3-sqrt(2)/2,0.5-sqrt(2)/2));
label("$F_2$", (4.5, 1.5), N);
label("$R$ = 1 m",(3, -0.5),S);
[/asy]

$ \textbf{(A)}\ 0.25\text{ N}\qquad\textbf{(B)}\ 0.5\text{ N}\qquad\textbf{(C)}\ 1\text{ N}\qquad\textbf{(D)}\ 2\text{ N}\qquad\textbf{(E)}\ 4\text{ N} $
1. **Identify the given quantities and the required relationship:**
   - Mass of each wheel, \( m = 1 \text{ kg} \)
   - Radius of the first wheel, \( R_1 = 0.5 \text{ m} \)
   - Radius of the second wheel, \( R_2 = 1 \text{ m} \)
   - Force applied on the first wheel, \( F_1 = 1 \text{ N} \)
   - Force applied on the second wheel, \( F_2 \) (to be determined)
   - Rotational inertia of a wheel, \( I = mR^2 \)
   - Angular acceleration, \( \alpha \)

2. **Use the relationship between torque, rotational inertia, and angular acceleration:**
   \[
   \tau = I \alpha
   \]
   where \( \tau \) is the torque, \( I \) is the rotational inertia, and \( \alpha \) is the angular acceleration.

3. **Calculate the torque for each wheel:**
   - For the first wheel:
     \[
     \tau_1 = F_1 R_1
     \]
   - For the second wheel:
     \[
     \tau_2 = F_2 R_2
     \]

4. **Express the rotational inertia for each wheel:**
   - For the first wheel:
     \[
     I_1 = m R_1^2 = 1 \cdot (0.5)^2 = 0.25 \text{ kg} \cdot \text{m}^2
     \]
   - For the second wheel:
     \[
     I_2 = m R_2^2 = 1 \cdot (1)^2 = 1 \text{ kg} \cdot \text{m}^2
     \]

5. **Set up the equation for angular acceleration:**
   Since the angular accelerations are identical (\(\alpha_1 = \alpha_2\)):
   \[
   \alpha = \frac{\tau_1}{I_1} = \frac{\tau_2}{I_2}
   \]

6. **Substitute the expressions for torque and rotational inertia:**
   \[
   \frac{F_1 R_1}{I_1} = \frac{F_2 R_2}{I_2}
   \]

7. **Simplify the equation:**
   \[
   \frac{F_1 R_1}{m R_1^2} = \frac{F_2 R_2}{m R_2^2}
   \]
   \[
   \frac{F_1}{R_1} = \frac{F_2}{R_2}
   \]

8. **Solve for \( F_2 \):**
   \[
   F_2 = F_1 \frac{R_2}{R_1}
   \]
   \[
   F_2 = 1 \text{ N} \cdot \frac{1 \text{ m}}{0.5 \text{ m}}
   \]
   \[
   F_2 = 2 \text{ N}
   \]

The final answer is \( \boxed{2 \text{ N}} \).