To calculate the angular velocity of the disk after $3.5$ rotations, we need to use the principles of rotational motion. Here's a step-by-step solution:

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Step 1: Moment of Inertia of the Disk

The moment of inertia ($I$) of a solid disk about its central axis is given by:
$I = \frac{1}{2} M R^2$
Where:

• $M = 4 \, \text{kg}$ (mass of the disk),
• $R = \frac{\text{Diameter}}{2} = \frac{180 \, \text{mm}}{2} = 0.09 \, \text{m}$.

Substitute the values:
$I = \frac{1}{2} \times 4 \times (0.09)^2 = \frac{1}{2} \times 4 \times 0.0081 = 0.0162 \, \text{kg·m}^2$

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Step 2: Torque Produced by the Force

Torque ($\tau$) is calculated using:
$\tau = F \cdot R$
Where:

• $F = 50 \, \text{N}$ (force),
• $R = 0.09 \, \text{m}$.

Substitute the values:
$\tau = 50 \times 0.09 = 4.5 \, \text{N·m}$

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Step 3: Angular Acceleration

Angular acceleration ($\alpha$) is related to torque and moment of inertia:
$\alpha = \frac{\tau}{I}$
Substitute the values:
$\alpha = \frac{4.5}{0.0162} = 277.78 \, \text{rad/s}^2$

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Step 4: Total Angle in Radians

One complete rotation corresponds to $2\pi \, \text{radians}$. For $3.5$ rotations:
$\theta = 3.5 \times 2\pi = 7\pi \, \text{radians}$

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Step 5: Angular Velocity

Using the equation of rotational kinematics:
$\omega^2 = \omega_0^2 + 2\alpha\theta$
Where:

• $\omega_0 = 0 \, \text{rad/s}$ (initial angular velocity),
• $\alpha = 277.78 \, \text{rad/s}^2$,
• $\theta = 7\pi$.

Substitute the values:
$\omega^2 = 0 + 2 \times 277.78 \times 7\pi$
$\omega^2 = 2 \times 277.78 \times 21.99 = 12217.29$
$\omega = \sqrt{12217.29} \approx 110.52 \, \text{rad/s}$

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Final Answer:

The angular velocity of the disk after $3.5$ rotations is approximately:
$\boxed{110.52 \, \text{rad/s}}$

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Follow-Up Questions

1. How would the angular velocity change if the force was doubled?
2. What is the angular momentum of the disk at this angular velocity?
3. How much work is done by the force during the $3.5$ rotations?
4. What is the linear velocity of a point on the rim of the disk at this angular velocity?
5. How long does it take for the disk to complete the $3.5$ rotations under this force?

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Tip: Always check the units carefully (e.g., converting diameter to radius in meters) to ensure accuracy in physics problems.