We are tasked with finding the angular velocity of a disk after a force is applied. Let’s break this into steps.

Given:

• Mass of the disk, $m$: $3 \, \text{kg}$
• Force applied, $F$: $69 \, \text{N}$
• Diameter of the disk, $d$: $167 \, \text{mm} = 0.167 \, \text{m}$
• Radius of the disk, $r$: $\frac{d}{2} = 0.0835 \, \text{m}$
• Revolutions turned, $n$: $7.6$
• Moment of inertia of the disk, $I$: $\frac{1}{2} m r^2 = \frac{1}{2} (3)(0.0835)^2$
• Convert revolutions to radians: $\theta = 7.6 \times 2\pi = 15.2\pi \, \text{radians}$

We are solving for the angular velocity $\omega$ in rad/s.

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Step 1: Calculate torque

The torque applied by the force $\tau$ is: $\tau = F \cdot r = 69 \cdot 0.0835 = 5.7615 \, \text{Nm}.$

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Step 2: Angular acceleration

The angular acceleration $\alpha$ is given by: $\alpha = \frac{\tau}{I}.$ Substitute $I = \frac{1}{2} m r^2$: $I = \frac{1}{2} (3) (0.0835)^2 = 0.010455375 \, \text{kg·m}^2.$ $\alpha = \frac{5.7615}{0.010455375} \approx 550.98 \, \text{rad/s}^2.$

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Step 3: Final angular velocity

Using the rotational kinematic equation: $\omega^2 = \omega_0^2 + 2\alpha \theta,$ where $\omega_0 = 0$ (initial angular velocity), we have: $\omega^2 = 0 + 2 (550.98)(15.2\pi).$ Simplify: $\omega^2 = 2 \cdot 550.98 \cdot 15.2 \cdot 3.14159 \approx 52577.14.$ $\omega = \sqrt{52577.14} \approx 229.34 \, \text{rad/s}.$

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Final Answer:

The angular velocity of the disk is approximately: $\boxed{229.34 \, \text{rad/s}}.$

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Do you need detailed clarifications or explanations for any of the steps?
Here are 5 related questions:

1. How does the torque formula change if the force is applied at an angle?
2. What happens if the disk is hollow instead of solid?
3. Can we calculate the time taken for the disk to reach this angular velocity?
4. How is this analysis different if the disk rolls without slipping?
5. How would the result change if the force increases?

Tip: Always check unit conversions (e.g., mm to meters) carefully to avoid calculation errors.