A cube hole is made in a solid hemisphere (radius R) such that maximum material is removed from the hemisphere. The centre of mass of the remaining body is at distance h from the centre of the circular base of the original hemisphere. Then h is
Question
A cube hole is made in a solid hemisphere (radius R) such that maximum material is removed from the hemisphere. The centre of mass of the remaining body is at distance h from the centre of the circular base of the original hemisphere. Then h is
Solution
The problem involves the concept of center of mass. The center of mass of a system of particles is the point that moves as though all of the system's mass were concentrated there and all external forces were applied there.
The center of mass of a hemisphere of radius R is located at a distance of 3R/8 from its base. When a cube of side a is removed from the hemisphere, the center of mass shifts due to the removal of the mass.
The maximum material removed from the hemisphere would be when the cube removed is inscribed in the hemisphere. The side of the cube, a, would be √2*R, and its center of mass would be at a distance of a/2 from the base of the hemisphere.
The new center of mass of the system (hemisphere - cube) can be calculated using the formula for the center of mass of two bodies:
h = (m1r1 - m2r2) / (m1 - m2)
where m1 and m2 are the masses of the hemisphere and cube respectively, and r1 and r2 are the distances of their centers of mass from the base of the hemisphere.
The mass of a body is proportional to its volume, so we can replace m1 and m2 with the volumes of the hemisphere and cube respectively. The volume of a hemisphere is 2/3πR^3 and the volume of a cube is a^3.
Substituting these values into the formula gives:
h = [(2/3πR^3)(3R/8) - (√2R)^3*(√2R/2)] / [(2/3πR^3) - (√2R)^3]
Solving this equation will give the value of h, the distance of the new center of mass from the base of the hemisphere.
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