A certain merry-go-round is accelerated uniformly from rest and attains an angular speed of 0.4 rad/s in the first 10 s. If the net applied torque is 2000 `N*m`, what is the moment of inertia of the merry-go-round?

Options

A. `400kg*m^(2)`
B. `800kg*m^(2)`
C. `50000kg*m^(2)`
D. `5000kg*m^(2)`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given data
- Initial angular velocity (\( \omega_0 \)) = 0 rad/s (since it starts from rest)
- Final angular velocity (\( \omega_f \)) = 0.4 rad/s
- Time (\( t \)) = 10 s
- Net applied torque (\( \tau \)) = 2000 N·m

### Step 2: Calculate the angular acceleration (\( \alpha \))
We can use the formula for angular acceleration:
\[ \alpha = \frac{\Delta \omega}{t} \]
where \( \Delta \omega = \omega_f - \omega_0 \).

Substituting the values:
\[ \Delta \omega = 0.4 \, \text{rad/s} - 0 \, \text{rad/s} = 0.4 \, \text{rad/s} \]
\[ \alpha = \frac{0.4 \, \text{rad/s}}{10 \, \text{s}} = 0.04 \, \text{rad/s}^2 \]

### Step 3: Use the torque equation to find the moment of inertia (\( I \))
The relationship between torque, moment of inertia, and angular acceleration is given by:
\[ \tau = I \cdot \alpha \]
Rearranging this formula to solve for \( I \):
\[ I = \frac{\tau}{\alpha} \]

### Step 4: Substitute the values into the equation
Substituting the values of torque and angular acceleration:
\[ I = \frac{2000 \, \text{N·m}}{0.04 \, \text{rad/s}^2} \]

### Step 5: Calculate the moment of inertia
Calculating the above expression:
\[ I = \frac{2000}{0.04} = 50000 \, \text{kg·m}^2 \]

### Conclusion
The moment of inertia of the merry-go-round is \( 50000 \, \text{kg·m}^2 \).

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