Final answer:
The moment of inertia for three 0.100 kg masses arranged at the corners of a square 0.25 m on each side is calculated to be approximately 0.00940 kg⋅m².
Step-by-step explanation:
To calculate the moment of inertia of three 0.100 kg masses arranged in a square configuration, we assume each mass is at the corner of a square with sides of 0.25 m. When calculating the moment of inertia about the center of the square, we can use the parallel-axis theorem and the properties of a rod with mass concentrated at its ends.
The moment of inertia (I) for a mass point is given by the formula I = mr², where m is the mass and r is the distance from the axis of rotation. Since the masses are at the corners of the square, the distance from the center for each mass is half the diagonal of the square (since the diagonal is the hypotenuse of a right-angle triangle formed by two adjacent sides of the square).
The length of the diagonal (d) can be found using Pythagoras' theorem:
d = √(0.25² + 0.25²) = √(0.0625 + 0.0625) = √(0.125) = 0.354 m approximately.
The distance r from the center to any corner is then d/2 = 0.354 m / 2 = 0.177 m.
Using the moment of inertia formula for point masses about the center, we get I = mr² for each mass. Here m = 0.100 kg and r = 0.177 m. Therefore, for one mass, I = 0.100 kg × (0.177 m)².
Summing up the contributions from all three masses, the total moment of inertia I_total = 3 × (0.100 kg × (0.177 m)²).
Perform the calculation:
3 × (0.100 kg × (0.177 m)²) = 3 × (0.100 kg × 0.031329) = 3 × 0.0031329 kg⋅m² = 0.0093987 kg⋅m².
Finally, the total moment of inertia for the three masses about the center of the square is approximately 0.00940 kg⋅m² (rounded to four decimal places).