Two wheels A and B are mounted on the same axle. Moment of inertia of A is 6 kg `m^(2)` and is rotated at 600 rpm, when B is at rest. What will be moment of inertia of B, if their combined speed is 400 rpm?

Options

A. `8 kg m^(2)`
B. `12 kg m^(2)`
C. `3 kg m^(2)`
D. `5 kg m^(2)`

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. Here's the step-by-step solution:

### Step 1: Understand the Problem
We have two wheels A and B mounted on the same axle. The moment of inertia of wheel A (I_A) is given as 6 kg·m², and it is rotating at 600 rpm. Wheel B is initially at rest. We need to find the moment of inertia of wheel B (I_B) when the combined speed of both wheels is 400 rpm.

### Step 2: Convert RPM to Radians per Second
First, we need to convert the rotational speeds from revolutions per minute (rpm) to radians per second (rad/s). The conversion factor is:
\[ \omega = 2\pi \times \left(\frac{n}{60}\right) \]
where \( n \) is the speed in rpm.

- For wheel A:
\[ \omega_A = 2\pi \times \left(\frac{600}{60}\right) = 20\pi \, \text{rad/s} \]

- For wheel B (initially at rest):
\[ \omega_B = 0 \, \text{rad/s} \]

- For the combined speed:
\[ \omega_{combined} = 2\pi \times \left(\frac{400}{60}\right) = \frac{40\pi}{6} \, \text{rad/s} \]

### Step 3: Apply the Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ I_A \omega_A + I_B \omega_B = (I_A + I_B) \omega_{combined} \]

Substituting the known values:
\[ 6 \cdot (20\pi) + I_B \cdot 0 = (6 + I_B) \cdot \left(\frac{40\pi}{6}\right) \]

### Step 4: Simplify the Equation
Since \( I_B \cdot 0 = 0 \), we can simplify the equation:
\[ 120\pi = (6 + I_B) \cdot \left(\frac{40\pi}{6}\right) \]

### Step 5: Cancel \( \pi \) and Solve for \( I_B \)
Dividing both sides by \( \pi \):
\[ 120 = (6 + I_B) \cdot \left(\frac{40}{6}\right) \]

Now, multiply both sides by \( \frac{6}{40} \):
\[ 120 \cdot \frac{6}{40} = 6 + I_B \]

Calculating the left side:
\[ 18 = 6 + I_B \]

### Step 6: Isolate \( I_B \)
Now, isolate \( I_B \):
\[ I_B = 18 - 6 = 12 \, \text{kg·m²} \]

### Final Answer
The moment of inertia of wheel B is:
\[ I_B = 12 \, \text{kg·m²} \]

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