$$F = I \alpha$$
Where F is the friction force, I is the moment of inertia of the disk, and $\alpha$ is the angular acceleration.
First, we need to calculate the moment of inertia of the disk. For a solid disk, the moment of inertia is given by:
$$I = \frac{1}{2} mR^2$$
Where m is the mass of the disk and R is the radius of the disk.
Substituting the given values, we get:
$$I = \frac{1}{2} \times 2.8kg \times (0.2m)^2 = 0.056kgm2$$
Next, we need to calculate the angular acceleration. The angular acceleration is given by:
$$\alpha = \frac{\Delta \omega}{\Delta t}$$
Where $\Delta \omega$ is the change in angular velocity and $\Delta t$ is the change in time.
The initial angular velocity of the disk is given by:
$$\omega_i = 260 \text{rpm} = 260 \times \frac{2\pi}{60} = 27.4rads^{-1}$$
The final angular velocity of the disk is zero.
Therefore, the change in angular velocity is:
$$\Delta \omega = \omega_f - \omega_i = 0 - 27.4rads^{-1} = -27.4rads^{-1}$$
The change in time is given as 2.0s.
Therefore, the angular acceleration is:
$$\alpha = \frac{-27.4rads^{-1}}{2.0s} = -13.7rads^{-2}$$
Finally, we can calculate the friction force required to bring the disk to a halt:
$$F = I \alpha = 0.056kgm2 \times -13.7rads^{-2} = -0.77N$$
Therefore, the brake must apply a friction force of 0.77N to the rim of the disk to bring it to a halt in 2.0s.
