A cord is wound over the rim of a flywheel of mass 20 kg and radius 25 cm. A mass `2.5` kg attached to the cord is allowed to fall under gravity. Calculate the angular acceleration of the flywheel

Options

A. `25"rad/s"^(2)`
B. `20"rad/s"^(2)`
C. `10"rad/s"^(2)`
D. `5"rad/s"^(2)`

Explanation

To solve the problem of finding the angular acceleration of the flywheel, we can follow these steps:

### Step 1: Identify the forces acting on the falling mass
The mass \( m = 2.5 \, \text{kg} \) is falling under the influence of gravity. The force acting on it due to gravity is given by:
\[ F_g = m \cdot g \]
where \( g = 9.81 \, \text{m/s}^2 \).

### Step 2: Calculate the gravitational force
Substituting the values:
\[ F_g = 2.5 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 24.525 \, \text{N} \]

### Step 3: Write the equation of motion for the falling mass
The net force acting on the mass when it is falling can be expressed as:
\[ F_{\text{net}} = F_g - T \]
where \( T \) is the tension in the cord. According to Newton's second law:
\[ F_{\text{net}} = m \cdot a \]
Thus, we have:
\[ m \cdot a = F_g - T \]

### Step 4: Relate linear acceleration to angular acceleration
The linear acceleration \( a \) of the falling mass is related to the angular acceleration \( \alpha \) of the flywheel by the equation:
\[ a = r \cdot \alpha \]
where \( r = 0.25 \, \text{m} \) (the radius of the flywheel).

### Step 5: Write the torque equation for the flywheel
The torque \( \tau \) acting on the flywheel due to the tension in the cord is given by:
\[ \tau = T \cdot r \]
According to Newton's second law for rotation:
\[ \tau = I \cdot \alpha \]
where \( I \) is the moment of inertia of the flywheel. For a solid disk, the moment of inertia is given by:
\[ I = \frac{1}{2} M r^2 \]
where \( M = 20 \, \text{kg} \) is the mass of the flywheel.

### Step 6: Calculate the moment of inertia
Substituting the values:
\[ I = \frac{1}{2} \cdot 20 \, \text{kg} \cdot (0.25 \, \text{m})^2 = \frac{1}{2} \cdot 20 \cdot 0.0625 = 0.625 \, \text{kg} \cdot \text{m}^2 \]

### Step 7: Substitute the torque equation
Now we can substitute the torque equation:
\[ T \cdot r = I \cdot \alpha \]
Substituting for \( I \):
\[ T \cdot 0.25 = 0.625 \cdot \alpha \]

### Step 8: Solve for tension \( T \)
From the equation of motion for the falling mass:
\[ 2.5 \cdot a = 24.525 - T \]
Substituting \( a = r \cdot \alpha = 0.25 \cdot \alpha \):
\[ 2.5 \cdot (0.25 \cdot \alpha) = 24.525 - T \]
This simplifies to:
\[ 0.625 \alpha = 24.525 - T \]
Thus, we can express \( T \) as:
\[ T = 24.525 - 0.625 \alpha \]

### Step 9: Substitute \( T \) back into the torque equation
Substituting this expression for \( T \) into the torque equation:
\[ (24.525 - 0.625 \alpha) \cdot 0.25 = 0.625 \cdot \alpha \]

### Step 10: Solve for angular acceleration \( \alpha \)
Expanding and rearranging gives:
\[ 6.13125 - 0.15625 \alpha = 0.625 \alpha \]
Combining terms:
\[ 6.13125 = 0.625 \alpha + 0.15625 \alpha \]
\[ 6.13125 = 0.78125 \alpha \]
Finally, solving for \( \alpha \):
\[ \alpha = \frac{6.13125}{0.78125} \approx 7.85 \, \text{rad/s}^2 \]

### Final Answer:
The angular acceleration of the flywheel is approximately \( 7.85 \, \text{rad/s}^2 \).

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