Question 1186952: Hello Good Day can you pls help me to solve my problem?
Four identical spheres of masses 2.0 kg each and radius 0.25 m are situated at the four corners of a square. One side of the square measures 3.00 m. Find the moment of inertia about an axis (a) passing through the center of the square and perpendicular to its plane, (b) passing through one of the masses and perpendicular to its plane, (c) passing through two adjacent masses parallel to the plane, and (d) in the plane running diagonally through two meters.
Thank you in advance❤️
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to calculate the moment of inertia for each scenario:
**Understanding Moment of Inertia**
Moment of inertia (I) is a measure of an object's resistance to rotational acceleration. It depends on the mass distribution of the object and the axis of rotation. For a point mass *m* rotating at a distance *r* from the axis, I = mr². For extended objects, we often use the parallel axis theorem: I = I_cm + md², where I_cm is the moment of inertia about the center of mass, *m* is the mass, and *d* is the distance between the axis of rotation and the center of mass.
**1. Moment of Inertia of a Single Sphere**
The moment of inertia of a solid sphere about an axis through its center is (2/5)mr². In our case:
I_sphere = (2/5) * 2.0 kg * (0.25 m)² = 0.05 kg*m²
**2. Calculations**
**(a) Axis through the center of the square and perpendicular to its plane:**
* Distance from each sphere to the axis: Half the diagonal of the square. The diagonal is √2 * side, so half the diagonal is (√2 * 3.00 m) / 2 = 2.12 m (approximately).
* Using the parallel axis theorem for each sphere: I = I_sphere + m * d² = 0.05 kg*m² + 2.0 kg * (2.12 m)² = 9.03 kg*m²
* Since there are four spheres: I_total = 4 * 9.03 kg*m² = 36.12 kg*m²
**(b) Axis through one of the masses and perpendicular to its plane:**
* One sphere is on the axis, so its contribution is just I_sphere.
* The other three spheres are at distances L, L, and √2 * L from the axis.
* I_total = I_sphere + m*(0)² + m*(3)² + m*(3)² + m*(3√2)² = 0.05 kg*m² + 2*9 kg*m² + 2*18 kg*m² = 72.05 kg*m² (approximately)
**(c) Axis through two adjacent masses parallel to the plane:**
* The two spheres on the axis contribute only their I_sphere terms.
* The other two spheres are at a distance of L from the axis.
* I_total = 2 * I_sphere + 2 * m * L² = 2 * 0.05 kg*m² + 2 * 2.0 kg * (3.00 m)² = 36.1 kg*m²
**(d) Axis in the plane running diagonally through two masses:**
* The two spheres on the axis contribute only their I_sphere terms.
* The other two spheres are at a distance of L/2 from the axis.
* I_total = 2 * I_sphere + 2 * m * (L/2)² = 2 * 0.05 kg*m² + 2 * 2.0 kg * (1.5 m)² = 9.1 kg*m²
**Summary of Answers (Approximate):**
* (a) 36.12 kg*m²
* (b) 72.05 kg*m²
* (c) 36.1 kg*m²
* (d) 9.1 kg*m²
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