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To solve for the rotational inertia (moment of inertia) \( I \) of the meter stick, we can treat it as a thin rod. The stick has a mass \( m = 0.289 \, \text{kg} \) and a length \( L = 1 \, \text{m} \). The axis of rotation is located at 16.8 cm from one end (or 0.168 m).

Since the axis is not at the center of mass, we can use the *parallel axis theorem*, which states:
\[
I = I_{\text{cm}} + m d^2
\]
where:
- \( I_{\text{cm}} \) is the moment of inertia about the center of mass,
- \( m \) is the mass of the rod,
- \( d \) is the distance between the center of mass and the new axis.

### Step 1: Calculate \( I_{\text{cm}} \)
For a thin rod rotating about its center, the moment of inertia \( I_{\text{cm}} \) is:
\[
I_{\text{cm}} = \frac{1}{12} m L^2
\]
Substitute \( m = 0.289 \, \text{kg} \) and \( L = 1 \, \text{m} \):
\[
I_{\text{cm}} = \frac{1}{12} \times 0.289 \times (1)^2 = \frac{0.289}{12} \approx 0.0241 \, \text{kg} \cdot \text{m}^2
\]

### Step 2: Calculate \( d \)
The distance \( d \) between the center of mass (which is at the 50 cm mark) and the axis (16.8 cm mark) is:
\[
d = 0.5 - 0.168 = 0.332 \, \text{m}
\]

### Step 3: Apply the Parallel Axis Theorem
Now, we can use the parallel axis theorem to find \( I \):
\[
I = I_{\text{cm}} + m d^2
\]
Substitute the values:
\[
I = 0.0241 + 0.289 \times (0.332)^2
\]
\[
I = 0.0241 + 0.289 \times 0.1102
\]
\[
I = 0.0241 + 0.0319 \approx 0.056 \, \text{kg} \cdot \text{m}^2
\]

### Answer
The rotational inertia of the stick about the given axis is approximately:
\[
\boxed{0.056 \, \text{kg} \cdot \text{m}^2}
\]

Would you like more details on the steps, or have any questions?

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**Further Questions for Practice:**
1. How would the moment of inertia change if the axis were at the midpoint of the stick?
2. How is the parallel axis theorem derived, and why is it applicable here?
3. What happens to the moment of inertia if the length of the stick is doubled?
4. How would we calculate the moment of inertia if the stick was not uniform in mass?
5. Can we apply a similar calculation for rotational inertia if the object is not a thin rod but a different shape?

**Tip:** Always remember that the parallel axis theorem is a powerful tool when dealing with objects rotated about an axis that is not through their center of mass.

Calculate the rotational inertia of a meter stick, with mass 0.289 kg, about an axis perpendicular to the stick and located at the 16.8 cm mark. (Treat the stick as a thin rod.)

Learn how to calculate the rotational inertia of a meter stick with a mass of 0.289 kg about an axis at 16.8 cm from one end. The solution involves using the parallel axis theorem and the moment of inertia formula for a thin rod. Suitable for advanced physics students.

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