A 20 kg square slab with side lengths 3.0 m rotates around an axis perpendicular to the slab. A particle of mass 0.60 kg sits on each of the corners of the square. Also, a particle of mass 0.80 kg sits at the center of each of the edges. What is the moment of inertia of the system?
Help me please :)
m1=20 kg, a=3 m,
I1 =m•a²/6 = 20•9/6 =30 kg•m²
m2 =0.6 kg , b =a•√2/2=3•√2/2 =2.13 m.
I2 =mb²=0.6•2.13²= 2.7 kg•m²
m3 = 0.8 kg, c=a/2 =1.5 m.
I3=mc²=0.8• 1.5² =1.8 kg•m².
I =I1+4•I2+4•I3 = 30+4•(2.7+1.8) =48 kg•m².
To find the moment of inertia of the system, we need to consider the contributions from each individual particle and combine them.
Let's break down the problem and calculate the moment of inertia of each component separately.
1. The four corner particles:
Each particle has a mass of 0.60 kg and is located at a distance of 1.5 m from the axis of rotation. The moment of inertia of each particle can be calculated using the formula:
I_corner = m * r^2
Where:
m = mass of the particle
r = distance from the axis of rotation
For each corner particle, the moment of inertia would be:
I_corner = (0.60 kg) * (1.5 m)^2 = 1.35 kg·m^2
Since we have four corner particles, the total moment of inertia contributed by the corner particles would be:
Total moment of inertia from corner particles = 4 * I_corner = 4 * 1.35 kg·m^2 = 5.40 kg·m^2
2. The four edge particles:
Each edge particle has a mass of 0.80 kg and is located at a distance of 1.5 m from the axis of rotation. However, the edge particles are closer to the axis compared to the corner particles. Hence, we need to apply the parallel axis theorem to calculate the moment of inertia.
The parallel axis theorem states that the moment of inertia of a particle about an axis parallel to and at a distance from the original axis is given by:
I_parallel = I_cm + (m * d^2)
Where:
I_cm = moment of inertia about the center of mass of the particle
m = mass of the particle
d = distance between the two axes
The moment of inertia about the center of mass of each edge particle is given by:
I_cm = (1/2) * m * r^2
Where:
m = mass of the particle
r = distance from the center of mass to the axis of rotation
For each edge particle, applying the parallel axis theorem, we have:
I_parallel = (1/2) * m * r^2 + (m * d^2)
Substituting the given values, the moment of inertia for each edge particle would be:
I_parallel = (1/2) * (0.80 kg) * (1.5 m)^2 + (0.80 kg) * (1.5 m)^2 = 1.80 kg·m^2 + 3.60 kg·m^2 = 5.40 kg·m^2
Since we have four edge particles, the total moment of inertia contributed by the edge particles would be:
Total moment of inertia from edge particles = 4 * I_parallel = 4 * 5.40 kg·m^2 = 21.60 kg·m^2
3. The slab itself:
The slab has a mass of 20 kg and is rotating as a whole. We can consider it as a uniform square plate rotating about its perpendicular axis.
The moment of inertia of a square plate rotating about its center and perpendicular to its plane is given by:
I_slab = (1/6) * m * a^2
Where:
m = mass of the plate
a = side length of the plate
Substituting the given values, we can calculate the moment of inertia of the slab:
I_slab = (1/6) * (20 kg) * (3.0 m)^2 = 30 kg·m^2
Finally, to get the total moment of inertia of the system, we sum the contributions from the corner particles, edge particles, and the slab:
Total moment of inertia = Total moment of inertia from corner particles + Total moment of inertia from edge particles + Moment of inertia of the slab
Total moment of inertia = 5.40 kg·m^2 + 21.60 kg·m^2 + 30 kg·m^2 = 57 kg·m^2
Therefore, the moment of inertia of the system is 57 kg·m^2.