A flywheel has moment of inertia `4kg-m^(2)` and has kinetic energy of 200 J. Calculate the number of revolutions it makes before coming to rest if a constant opposing couple of 5N-m is applied to the flywheel

Options

A. 12.8 rev
B. 24 rev
C. 6.4 rev
D. 16 rev

Answer

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

### Step 1: Identify the given values
- Moment of inertia (I) = 4 kg·m²
- Kinetic energy (KE) = 200 J
- Opposing couple (τ) = 5 N·m

### Step 2: Relate kinetic energy to angular velocity
The kinetic energy of a rotating object is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
We can rearrange this to find the angular velocity (ω):
\[ \omega^2 = \frac{2 \cdot KE}{I} \]
Substituting the values:
\[ \omega^2 = \frac{2 \cdot 200 \, \text{J}}{4 \, \text{kg·m}^2} = \frac{400}{4} = 100 \]
Taking the square root:
\[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]

### Step 3: Calculate angular acceleration (α)
The angular acceleration can be found using the formula:
\[ \tau = I \alpha \]
Rearranging gives:
\[ \alpha = \frac{\tau}{I} \]
Substituting the values:
\[ \alpha = \frac{-5 \, \text{N·m}}{4 \, \text{kg·m}^2} = -\frac{5}{4} \, \text{rad/s}^2 = -1.25 \, \text{rad/s}^2 \]

### Step 4: Use the rotational motion equation to find θ
We will use the equation:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Where:
- \(\omega = 0\) (final angular velocity, as the flywheel comes to rest)
- \(\omega_0 = 10 \, \text{rad/s}\) (initial angular velocity)
- \(\alpha = -1.25 \, \text{rad/s}^2\)

Substituting the values:
\[ 0 = (10)^2 + 2(-1.25)\theta \]
This simplifies to:
\[ 0 = 100 - 2.5\theta \]
Rearranging gives:
\[ 2.5\theta = 100 \]
Thus:
\[ \theta = \frac{100}{2.5} = 40 \, \text{radians} \]

### Step 5: Convert θ to number of revolutions (n)
The number of revolutions can be calculated using the relation:
\[ n = \frac{\theta}{2\pi} \]
Substituting the value of θ:
\[ n = \frac{40}{2\pi} = \frac{40}{6.2832} \approx 6.37 \]

### Final Answer
The number of revolutions the flywheel makes before coming to rest is approximately **6.37 revolutions**.

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