A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from `6 kg-m^2 to 2 kg-m^2` what will be the new angular speed?

Explanation

To find the new angular speed of the system after the boy pulls the balls closer, we can use the principle of conservation of angular momentum. Here’s a step-by-step solution:

### Step 1: Understand the conservation of angular momentum
The angular momentum \( L \) of a system remains constant if no external torque acts on it. The angular momentum can be expressed as:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular speed.

### Step 2: Set up the equation for initial and final states
According to the conservation of angular momentum:
\[ I_1 \cdot \omega_1 = I_2 \cdot \omega_2 \]
where:
- \( I_1 = 6 \, \text{kg-m}^2 \) (initial moment of inertia)
- \( \omega_1 = 120 \, \text{revolutions per minute} \)
- \( I_2 = 2 \, \text{kg-m}^2 \) (final moment of inertia)
- \( \omega_2 \) is the final angular speed we need to find.

### Step 3: Convert angular speed from revolutions per minute to radians per second
To use consistent units, we convert \( \omega_1 \) from revolutions per minute (rpm) to radians per second:
\[ \omega_1 = 120 \, \text{rpm} = 120 \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = 12.5664 \, \text{radians/second} \]

### Step 4: Substitute values into the conservation equation
Now, substituting the known values into the equation:
\[ 6 \cdot 12.5664 = 2 \cdot \omega_2 \]

### Step 5: Solve for \( \omega_2 \)
Now, we can solve for \( \omega_2 \):
\[ 75.3984 = 2 \cdot \omega_2 \]
\[ \omega_2 = \frac{75.3984}{2} = 37.6992 \, \text{radians/second} \]

### Step 6: Convert back to revolutions per minute
Finally, convert \( \omega_2 \) back to revolutions per minute:
\[ \omega_2 = 37.6992 \times \frac{1 \, \text{revolution}}{2\pi \, \text{radians}} \times 60 \, \text{seconds} \approx 360 \, \text{rpm} \]

### Final Answer
The new angular speed of the system is:
\[ \omega_2 \approx 360 \, \text{revolutions per minute} \]

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