Question

A hollow sphere of mass 500 gm and radius 50 cm, is joined at the bottom of a rod of mass 1 kg and length 3m. Find the moment of inertia of the system about the axis passing through the topmost point of the rod and perpendicular to the rod.

Answer 1

$\frac{91}{12}$

Answer 2

The system consists of a rod and a hollow sphere. The axis of rotation passes through the topmost point of the rod and is perpendicular to the rod. The total moment of inertia of the system is the sum of the moment of inertia of the rod and the moment of inertia of the sphere about this axis.

1. Moment of inertia of the rod: The rod has mass $M_{rod} = 1$ kg and length $L = 3$ m. The axis of rotation passes through one end of the rod (the topmost point) and is perpendicular to its length. The moment of inertia of a rod about an axis perpendicular to its length and passing through one end is given by the formula: $I_{rod} = \frac{1}{3} M_{rod} L^2$ Substituting the given values: $I_{rod} = \frac{1}{3} (1 \text{ kg}) (3 \text{ m})^2 = \frac{1}{3} \times 1 \times 9 = 3 \text{ kg m}^2$.

2. Moment of inertia of the hollow sphere: The hollow sphere has mass $M_{sphere} = 500$ gm $= 0.5$ kg and radius $R = 50$ cm $= 0.5$ m. The sphere is joined at the bottom of the rod. We assume that "joined at the bottom of a rod" means the center of the sphere is located at the bottom end of the rod. The distance of the bottom end of the rod from the topmost point is $L = 3$ m. So, the distance of the center of the sphere from the axis of rotation is $d = L = 3$ m. The moment of inertia of a hollow sphere about an axis passing through its center of mass is $I_{CM} = \frac{2}{3} M_{sphere} R^2$. $I_{CM} = \frac{2}{3} (0.5 \text{ kg}) (0.5 \text{ m})^2 = \frac{2}{3} \times 0.5 \times 0.25 = \frac{1}{3} \times 0.25 = \frac{0.25}{3} = \frac{1}{12} \text{ kg m}^2$. Since the axis of rotation does not pass through the center of mass of the sphere, we use the parallel axis theorem. The moment of inertia of the sphere about the given axis is: $I_{sphere} = I_{CM} + M_{sphere} d^2$ $I_{sphere} = \frac{1}{12} + (0.5 \text{ kg}) (3 \text{ m})^2 = \frac{1}{12} + 0.5 \times 9 = \frac{1}{12} + 4.5$. To add these values, we convert 4.5 to a fraction: $4.5 = \frac{9}{2}$. $I_{sphere} = \frac{1}{12} + \frac{9}{2} = \frac{1}{12} + \frac{9 \times 6}{2 \times 6} = \frac{1}{12} + \frac{54}{12} = \frac{1 + 54}{12} = \frac{55}{12} \text{ kg m}^2$.

3. Total moment of inertia of the system: The total moment of inertia is the sum of the moments of inertia of the rod and the sphere. $I_{system} = I_{rod} + I_{sphere}$ $I_{system} = 3 \text{ kg m}^2 + \frac{55}{12} \text{ kg m}^2 = \frac{3 \times 12}{12} + \frac{55}{12} = \frac{36}{12} + \frac{55}{12} = \frac{36 + 55}{12} = \frac{91}{12} \text{ kg m}^2$.

The moment of inertia of the system about the axis passing through the topmost point of the rod and perpendicular to the rod is $\frac{91}{12}$ kg m$^2$.