A 50 N m torque acts on a wheel with a moment of inertia `150kg*m^(2)`. If the wheel starts from rest, how long will it take the wheel to make 1 rev?

Options

A. 2.4 s
B. 10 s
C. 0.66 s
D. 6.1 s

Answer

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

### Step 1: Identify the given values
- Torque (τ) = 50 N·m
- Moment of inertia (I) = 150 kg·m²
- Initial angular velocity (ω₀) = 0 (since the wheel starts from rest)

### Step 2: Calculate the angular acceleration (α)
Using the formula for torque:
\[ \tau = I \cdot \alpha \]
We can rearrange this to find angular acceleration (α):
\[ \alpha = \frac{\tau}{I} \]
Substituting the known values:
\[ \alpha = \frac{50 \, \text{N·m}}{150 \, \text{kg·m²}} = \frac{1}{3} \, \text{rad/s²} \]

### Step 3: Determine the angular displacement (θ) for one revolution
One complete revolution corresponds to an angular displacement of:
\[ \theta = 2\pi \, \text{radians} \]

### Step 4: Use the angular kinematics equation
Since the wheel starts from rest, we can use the following kinematic equation for rotational motion:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
Since the initial angular velocity (ω₀) is 0, the equation simplifies to:
\[ \theta = \frac{1}{2} \alpha t^2 \]
Substituting the values we have:
\[ 2\pi = \frac{1}{2} \cdot \frac{1}{3} \cdot t^2 \]

### Step 5: Solve for time (t)
Rearranging the equation:
\[ 2\pi = \frac{1}{6} t^2 \]
Multiplying both sides by 6:
\[ 12\pi = t^2 \]
Taking the square root of both sides:
\[ t = \sqrt{12\pi} \]

### Step 6: Calculate the numerical value
Using the approximate value of π (3.14):
\[ t = \sqrt{12 \cdot 3.14} \approx \sqrt{37.68} \approx 6.14 \, \text{seconds} \]

### Final Answer
The time it will take for the wheel to make one revolution is approximately **6.14 seconds**.

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