Two identical spheres each of mass 1.20 kg and radius 10.0 cm are fixed at the ends of a light rod so that the separation between the centres is 50.0 cm. Find the moment of intertia of the system about an axis perpendicular to the rod passing through its middle point.
Two identical spheres each of mass 1.20 kg and radius 10.0 cm are fixed at the ends of a light rod so that the separation between the centres is 50.0 cm. Find the moment of intertia of the system about an axis perpendicular to the rod passing through its middle point.
The correct option is D 0.16kg - m2
Consider the diameter of one of the spheres parallel to the given axis. The moment of Inertia of this sphere about the diameter is
I = 25mR2 = 25(1.20 kg)(0.1 m)2
= 4.8 × 10−3 kg−m2.
Its moment of inertia about the given axis is obtained by using the parallel axes theorem. Thus,
I = Icm + md2
= 4.8 × 10−3 kg−m2 + (1.20 kg)(0.25m)2
= 4.8 × 10−3 kg−m2 + 0.075 kg−m2
= 79.8 × 10−3 kg−m2.
The amount of inertia of the second sphere is also the same, so that the moment of inertia of the system is
2 × 79.8 × 10−3 kg−m2 ≈ 0.160 kg−m2.
0.16kg - m2
Consider the diameter of one of the spheres parallel to the given axis. The moment of Inertia of this sphere about the diameter is
I = 25mR2 = 25(1.20 kg)(0.1 m)2
= 4.8 × 10−3 kg−m2.
Its moment of inertia about the given axis is obtained by using the parallel axes theorem. Thus,
I = Icm + md2
= 4.8 × 10−3 kg−m2 + (1.20 kg)(0.25m)2
= 4.8 × 10−3 kg−m2 + 0.075 kg−m2
= 79.8 × 10−3 kg−m2.
The amount of inertia of the second sphere is also the same, so that the moment of inertia of the system is
2 × 79.8 × 10−3 kg−m2 ≈ 0.160 kg−m2.
