The center of mass is L/3 from mass 2m and 2L/3 from mass m. Add the moment of inertia contributions of each mass to get
m*(2L/3)^2 + (2m)(L/3)^2
= mL^2*(4/9 + 2/9)
= (2/3) mL^2
m*(2L/3)^2 + (2m)(L/3)^2
= mL^2*(4/9 + 2/9)
= (2/3) mL^2
The moment of inertia of an object depends on its mass and distribution of mass around the axis of rotation. For a point mass rotating about an axis at a distance 'r' from the axis, the moment of inertia is given by:
I = m * r^2
Let's consider the two objects individually:
1. Object with mass 'm':
Since the object is attached at the end of the rod, its distance from the axis of rotation is half the length of the rod (L/2). Therefore, its moment of inertia is:
I1 = m * (L/2)^2
2. Object with mass '2m':
Similarly, the distance of this object from the axis of rotation is also L/2. So, its moment of inertia is:
I2 = (2m) * (L/2)^2
Now, to find the total moment of inertia of the arrangement, we can add the individual moments of inertia together:
Total moment of inertia = I1 + I2
Simplifying it:
Total moment of inertia = m * (L/2)^2 + (2m) * (L/2)^2
= [m + 4m] * (L/2)^2
= 5m * (L/2)^2
= 5m * L^2 / 4
Therefore, the moment of inertia of the arrangement about an axis through the center of mass, perpendicular to the rod, is 5m * L^2 / 4.