To solve this problem, let’s break down the information provided:

Given Data:

1. Radius of the disk: $R = 0.400 \, \text{m}$
2. Force applied: $F = 6.00 \, \text{N}$
3. Time duration: $t = 4.00 \, \text{s}$
4. Final angular velocity: $\omega = 12.0 \, \text{rad/s}$
5. Moment of inertia of a uniform disk: $I = \frac{1}{2} m R^2$

Steps to Solve:

1. Calculate the Torque ($\tau$):

   • Torque is given by $\tau = F \cdot R$.
   • Substitute $F = 6.00 \, \text{N}$ and $R = 0.400 \, \text{m}$: $\tau = 6.00 \times 0.400 = 2.4 \, \text{Nm}$

2. Calculate Angular Acceleration ($\alpha$):

   • Torque is also related to angular acceleration by $\tau = I \alpha$.
   • Substitute $I = \frac{1}{2} m R^2$: $\tau = \frac{1}{2} m R^2 \cdot \alpha$
   • Solving for $\alpha$: $\alpha = \frac{2 \tau}{m R^2}$

3. Relate Angular Acceleration and Final Angular Velocity:

   • Using $\omega = \alpha t$ to find $\alpha$: $\alpha = \frac{\omega}{t} = \frac{12.0}{4.0} = 3.0 \, \text{rad/s}^2$

4. Solve for the Mass $m$:

   • Substitute $\alpha = 3.0 \, \text{rad/s}^2$ and $\tau = 2.4 \, \text{Nm}$: $2.4 = \frac{1}{2} m (0.4)^2 \cdot 3.0$ $m = \frac{2.4}{0.24} = 10.0 \, \text{kg}$

Answer:

The mass of the disk is 10.0 kg (Option C).

Would you like further explanation on any of the steps or have additional questions?

Additional Questions:

1. What is the relationship between torque and angular acceleration?
2. How would the mass change if the applied force were different?
3. What would be the effect of a larger radius on the angular acceleration?
4. How does the time duration affect the final angular velocity?
5. Why is the moment of inertia different for various shapes?

Tip:

Remember, torque and force are related but act differently in rotational and linear dynamics.