To solve this problem, we can divide it into two parts:

1. The first 5 seconds (when the wheel is accelerating).
2. The next 5 seconds (when the wheel moves at a constant angular velocity).

Step 1: The first 5 seconds (with angular acceleration)

Let:

• $\alpha$ be the angular acceleration (constant).
• $\omega_0$ be the initial angular velocity.
• $\theta_1$ be the angular displacement during the first 5 seconds, which is given as $100 \, \text{radians}$.
• $t_1 = 5 \, \text{seconds}$ is the time for this period.

The equation for angular displacement when there's angular acceleration is: $\theta_1 = \omega_0 t_1 + \frac{1}{2} \alpha t_1^2$

Substituting the values: $100 = \omega_0 (5) + \frac{1}{2} \alpha (5^2)$ $100 = 5 \omega_0 + \frac{25}{2} \alpha \quad \text{(Equation 1)}$

Step 2: The next 5 seconds (with constant angular velocity)

During the second 5 seconds, the angular velocity is constant, meaning that the angular velocity at the end of the first 5 seconds is maintained.

Let $\omega_f$ be the final angular velocity at the end of the first 5 seconds. The angular displacement $\theta_2$ during the second 5 seconds is: $\theta_2 = \omega_f t_2$ where:

• $\theta_2 = 80 \, \text{radians}$,
• $t_2 = 5 \, \text{seconds}$.

Thus: $80 = \omega_f (5)$ $\omega_f = \frac{80}{5} = 16 \, \text{rad/s} \quad \text{(Equation 2)}$

Step 3: Relationship between $\omega_f$, $\omega_0$, and $\alpha$

The angular velocity at the end of the first 5 seconds is related to the initial angular velocity and the angular acceleration by: $\omega_f = \omega_0 + \alpha t_1$

Substitute $\omega_f = 16 \, \text{rad/s}$ and $t_1 = 5 \, \text{seconds}$: $16 = \omega_0 + 5\alpha \quad \text{(Equation 3)}$

Step 4: Solve the system of equations

Now, we have two equations:

1. $100 = 5\omega_0 + \frac{25}{2}\alpha$
2. $16 = \omega_0 + 5\alpha$

We can solve these equations simultaneously to find $\omega_0$ and $\alpha$.

From Equation 3, solve for $\omega_0$: $\omega_0 = 16 - 5\alpha$ Substitute this into Equation 1: $100 = 5(16 - 5\alpha) + \frac{25}{2} \alpha$ $100 = 80 - 25\alpha + \frac{25}{2} \alpha$ Multiply through by 2 to eliminate the fraction: $200 = 160 - 50\alpha + 25\alpha$ $200 = 160 - 25\alpha$ $40 = 25\alpha$ $\alpha = \frac{40}{25} = 1.6 \, \text{rad/s}^2$

Step 5: Find $\omega_0$

Now substitute $\alpha = 1.6 \, \text{rad/s}^2$ into Equation 3: $16 = \omega_0 + 5(1.6)$ $16 = \omega_0 + 8$ $\omega_0 = 16 - 8 = 8 \, \text{rad/s}$

Final Answers:

• The initial angular velocity, $\omega_0 = 8 \, \text{rad/s}$.
• The angular acceleration, $\alpha = 1.6 \, \text{rad/s}^2$.

Let me know if you would like any more details or if you have any questions.

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Here are some related questions:

1. What is the final angular velocity if the wheel continues accelerating for another 5 seconds?
2. How much angular displacement would the wheel describe after 10 seconds of constant acceleration?
3. What would happen if the angular acceleration was doubled for the first 5 seconds?
4. How would the results change if the constant angular velocity period was extended to 10 seconds instead of 5?
5. How can the moment of inertia of the wheel be calculated if the torque is known?

Tip: To solve problems involving rotational motion, remember that the equations mirror linear motion but replace displacement with angular displacement, velocity with angular velocity, and acceleration with angular acceleration.