two small objects of masses 'm' & '2m' each are attaached to the ends of a light rigid rod of length 'L' . the rod is pivoted at a point where centre of mass of the two objects lie. the moment of inertia of the arrangement about an axis through the centre of mass , perpendicularly to the rod is ..........
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
To find the moment of inertia of the arrangement about an axis through the center of mass, perpendicular to the rod, we need to consider the individual contributions of each object.
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.