A wheel of moment of inertia `0.10 kg-m^2` is rotating about a shaft at an angular speed of `160 (rev)/(min)`. A second wheel is set into rotation at 300 rev/minute and is coupled to the same shaft so that both the wheels finally rotate with as common angular speed of 200 rev/minute. Find the moment of inertia of the second wheel.

Explanation

To find the moment of inertia of the second wheel, we can use the principle of conservation of angular momentum. Here are the steps to solve the problem:

### Step 1: Identify the given values
- Moment of inertia of the first wheel, \( I_1 = 0.10 \, \text{kg m}^2 \)
- Angular speed of the first wheel, \( \omega_1 = 160 \, \text{rev/min} \)
- Angular speed of the second wheel, \( \omega_2 = 300 \, \text{rev/min} \)
- Final common angular speed, \( \omega = 200 \, \text{rev/min} \)

### Step 2: Convert angular speeds from revolutions per minute to radians per second
To use the angular speeds in calculations, we need to convert them to radians per second:
\[ \omega_1 = 160 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{160 \times 2\pi}{60} \approx 16.76 \, \text{rad/s} \]
\[ \omega_2 = 300 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{300 \times 2\pi}{60} \approx 31.42 \, \text{rad/s} \]
\[ \omega = 200 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{200 \times 2\pi}{60} \approx 20.94 \, \text{rad/s} \]

### Step 3: Apply the conservation of angular momentum
According to the conservation of angular momentum:
\[ I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega \]
Substituting the known values:
\[ 0.10 \times 16.76 + I_2 \times 31.42 = (0.10 + I_2) \times 20.94 \]

### Step 4: Simplify the equation
Calculating the left-hand side:
\[ 1.676 + I_2 \times 31.42 = (0.10 + I_2) \times 20.94 \]
Expanding the right-hand side:
\[ 1.676 + I_2 \times 31.42 = 0.10 \times 20.94 + I_2 \times 20.94 \]
\[ 1.676 + I_2 \times 31.42 = 2.094 + I_2 \times 20.94 \]

### Step 5: Rearranging the equation
Now, rearranging the equation gives:
\[ I_2 \times 31.42 - I_2 \times 20.94 = 2.094 - 1.676 \]
\[ I_2 \times (31.42 - 20.94) = 0.418 \]
\[ I_2 \times 10.48 = 0.418 \]

### Step 6: Solve for \( I_2 \)
Now, divide both sides by \( 10.48 \):
\[ I_2 = \frac{0.418}{10.48} \approx 0.0399 \, \text{kg m}^2 \]
Rounding off gives:
\[ I_2 \approx 0.04 \, \text{kg m}^2 \]

### Final Answer
The moment of inertia of the second wheel is approximately \( 0.04 \, \text{kg m}^2 \).

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