A flywheel of moment of inertia `10^(7) g cm^(2)` is rotating at a speed of 120 rotations per minute. Find the constant breaking torque required to stop the wheel in 5 rotations.

Answer

To solve the problem step by step, we will follow the concepts of rotational motion, including moment of inertia, angular velocity, angular displacement, and angular acceleration.

### Step 1: Convert the rotational speed from RPM to radians per second
The flywheel is rotating at a speed of 120 rotations per minute (RPM). We need to convert this to radians per second.

\[ \text{Angular velocity} (\omega) = 120 \, \text{RPM} \times \frac{2\pi \, \text{radians}}{1 \, \text{rotation}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} \]

Calculating this gives:

\[ \omega = 120 \times \frac{2\pi}{60} = 4\pi \, \text{radians/second} \]

### Step 2: Identify the parameters
- Moment of inertia \( I = 10^7 \, \text{g cm}^2 \)
- Initial angular velocity \( \omega_0 = 4\pi \, \text{radians/second} \)
- Final angular velocity \( \omega_f = 0 \, \text{radians/second} \) (since the wheel stops)
- Angular displacement \( \theta = 5 \, \text{rotations} = 5 \times 2\pi = 10\pi \, \text{radians} \)

### Step 3: Use the equation of motion for rotation
We will use the equation of motion for rotational dynamics:

\[ \omega_f^2 = \omega_0^2 + 2\alpha\theta \]

Substituting the known values:

\[ 0 = (4\pi)^2 + 2\alpha(10\pi) \]

This simplifies to:

\[ 0 = 16\pi^2 + 20\pi\alpha \]

### Step 4: Solve for angular acceleration \( \alpha \)
Rearranging the equation to solve for \( \alpha \):

\[ 20\pi\alpha = -16\pi^2 \]

Dividing both sides by \( 20\pi \):

\[ \alpha = -\frac{16\pi^2}{20\pi} = -\frac{16\pi}{20} = -\frac{4\pi}{5} \, \text{radians/second}^2 \]

### Step 5: Calculate the braking torque
The torque \( \tau \) required to produce this angular acceleration can be calculated using the formula:

\[ \tau = I \alpha \]

Substituting the values we have:

\[ \tau = 10^7 \, \text{g cm}^2 \times \left(-\frac{4\pi}{5} \, \text{radians/second}^2\right) \]

Calculating this gives:

\[ \tau = -\frac{4\pi \times 10^7}{5} \, \text{g cm}^2/\text{s}^2 \]

### Final Answer
The magnitude of the braking torque required is:

\[ \tau = \frac{4\pi \times 10^7}{5} \, \text{g cm}^2/\text{s}^2 \]