To find the time $t_2$ it will take for the wheel to stop after the brake is applied, we can use the following kinematic equation for rotational motion:

$\omega_f = \omega_i + \alpha t$

where:

• $\omega_f$ is the final angular velocity (which will be zero since the wheel is stopping),
• $\omega_i = \omega_1$ is the initial angular velocity at the moment the brake is applied,
• $\alpha = -5\alpha$ is the constant angular deceleration applied by the brake, and
• $t = t_2$ is the time we need to find.

Since we know that the wheel's angular velocity will eventually reach zero, we can set up the equation as:

$0 = \omega_1 + (-5\alpha)t_2$

Rearranging this equation to solve for $t_2$, we get:

$t_2 = \frac{\omega_1}{5\alpha}$

Thus, the time $t_2$ it will take the wheel to stop after the brake is applied is:

$t_2 = \frac{\omega_1}{5\alpha}$

This time depends on the initial angular velocity $\omega_1$ and the deceleration $-5\alpha$.

Would you like further details on how to apply this formula in specific cases or explanations of the variables?

Here are five related questions to explore the concept further:

1. How would $t_2$ change if the deceleration were doubled to $-10\alpha$?
2. What would be the effect on $t_2$ if $\omega_1$ were halved?
3. How would you calculate the angular displacement of the wheel during deceleration?
4. How does constant torque influence the angular acceleration in rotational motion?
5. If the flywheel started with an initial angular velocity, how would the approach change?

Tip: When solving problems involving constant acceleration or deceleration, kinematic equations for linear motion have direct analogs in rotational motion, where linear quantities (like velocity and acceleration) are replaced by angular quantities.