A disc at rest of Mars 100 kg and radius 0.4 m acquire a constant angular alciliation of eight radius calculate the angular velocity in two seconds calculate angular displacement calculate linear velocity calculate tangential acceleration calculate centripetal acceleration calculate resultant acceleration and calculate centripetal Force

The initial angular velocity is $\omega_0 = 0$ rad/s.
The constant angular acceleration is $\alpha = 8$ rad/s$^2$.
The time is $t = 2$ s.

Using the kinematic equation for angular velocity:

$$\omega = \omega_0 + \alpha t$$ $$\omega = 0 \ \mathrm{rad/s} + (8 \ \mathrm{rad/s^2})(2 \ \mathrm{s})$$ $$\omega = 16 \ \mathrm{rad/s}$$

The angular velocity in two seconds is $\boxed{16 \ \mathrm{rad/s}}$.

Using the kinematic equation for angular displacement:

$$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$$ $$\theta = (0 \ \mathrm{rad/s})(2 \ \mathrm{s}) + \frac{1}{2} (8 \ \mathrm{rad/s^2})(2 \ \mathrm{s})^2$$ $$\theta = 0 + \frac{1}{2} (8 \ \mathrm{rad/s^2})(4 \ \mathrm{s^2})$$ $$\theta = 16 \ \mathrm{rad}$$

The angular displacement is $\boxed{16 \ \mathrm{rad}}$.

The radius of the disc is $R = 0.4$ m.
The angular velocity at $t=2$ s is $\omega = 16$ rad/s (from Step 1).

Using the relation between linear and angular velocity:

$$v = R \omega$$ $$v = (0.4 \ \mathrm{m})(16 \ \mathrm{rad/s})$$ $$v = 6.4 \ \mathrm{m/s}$$

The linear velocity is $\boxed{6.4 \ \mathrm{m/s}}$.

The radius of the disc is $R = 0.4$ m.
The constant angular acceleration is $\alpha = 8$ rad/s$^2$.

Using the relation between tangential and angular acceleration:

$$a_t = R \alpha$$ $$a_t = (0.4 \ \mathrm{m})(8 \ \mathrm{rad/s^2})$$ $$a_t = 3.2 \ \mathrm{m/s^2}$$

The tangential acceleration is $\boxed{3.2 \ \mathrm{m/s^2}}$.

The radius of the disc is $R = 0.4$ m.
The angular velocity at $t=2$ s is $\omega = 16$ rad/s (from Step 1).

Using the formula for centripetal acceleration:

$$a_c = R \omega^2$$ $$a_c = (0.4 \ \mathrm{m})(16 \ \mathrm{rad/s})^2$$ $$a_c = (0.4 \ \mathrm{m})(256 \ \mathrm{rad^2/s^2})$$ $$a_c = 102.4 \ \mathrm{m/s^2}$$

The centripetal acceleration is $\boxed{102.4 \ \mathrm{m/s^2}}$.

The tangential acceleration is $a_t = 3.2$ m/s$^2$ (from Step 4).
The centripetal acceleration is $a_c = 102.4$ m/s$^2$ (from Step 5).

The resultant acceleration is the vector sum of the tangential and centripetal accelerations, which are perpendicular to each other:

$$a_{res} = \sqrt{a_t^2 + a_c^2}$$ $$a_{res} = \sqrt{(3.2 \ \mathrm{m/s^2})^2 + (102.4 \ \mathrm{m/s^2})^2}$$ $$a_{res} = \sqrt{10.24 \ \mathrm{m^2/s^4} + 10485.76 \ \mathrm{m^2/s^4}}$$ $$a_{res} = \sqrt{10496 \ \mathrm{m^2/s^4}}$$ $$a_{res} \approx 102.45 \ \mathrm{m/s^2}$$

The resultant acceleration is $\boxed{102.45 \ \mathrm{m/s^2}}$.

The mass of the disc is $m = 100$ kg.
The centripetal acceleration is $a_c = 102.4$ m/s$^2$ (from Step 5).

Using Newton's second law for centripetal force:

$$F_c = m a_c$$ $$F_c = (100 \ \mathrm{kg})(102.4 \ \mathrm{m/s^2})$$ $$F_c = 10240 \ \mathrm{N}$$

The centripetal force is $\boxed{10240 \ \mathrm{N}}$.