Ahollow sphere is attached to the end of a uniform rod. the sphere has a radius of 0.60 m and a mass of 0.44 kg. the rod has a length of 1.28 m and a mass of 0.48 kg. the rod is placed on a fulcrum (pivot) at x = 0.46 m from the left end of the rod. calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum.

Moment of inertia is the property of a body due to which it resists angular acceleration. It is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.

Explanation: I = m*r*r

Given mass of sphere = 0.4kg

Radius of sphere = 0.6m

Moment of Inertia of sphere = (2/3)*0.4* 0.6

Inertia of sphere = 0.096 kgm2

Given mass of rod = 0.48kg

Length of rod = 1.28m

Inertia of rod = (m*L*L)/3

I = (0.48 * 1.28 * 1.28)/3

I = 0.218 kgm2

Total Moment of inertia I = 0.096 + 0.218

I = 0.314 kgm2

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

their crystal patterns have been disrupted by stress