Question

An abstract sculpture consists of a ball (radius R = 76 cm) resting on top of a cube (each side L = 200 cm long). The ball and the cube are made of the same material of uniform density; there are no hollow spaces inside them. The bottom face of the cube rests on a horizontal floor. How high is the sculpture’s center of mass above the floor? Answer in units of cm.

Answer 1

Answer:

132.9 cm

Explanation:

Data provided in the question:

Radius of the basll = 76 cm = 0.76 m

Side of the box = 200 cm = 2 m

Density of the ball and cube are equal

let the density be 'D'

Now,

Mass of ball, M = Volume × Density

= $\frac{4}{3}\pi r^3$  × D

= $\frac{4}{3}\pi (0.76)^3$× D

= 1.838D

Mass of cube, m = L³ × D

= 2³ × D

= 8D

Thus,

center of mass, y = [ My₁ + my₂ ] ÷ [M + m]

here,

y₁ = center of mass of ball with respect to floor

as the center mass of sphere lies at the center of the sphere

= Length of cube + radius of sphere

= 2 + 0.76

= 2.76 m

y₂ = Center of mass of cube = $\frac{L}{2}=\frac{2}{2}$ = 1 m

Thus,

y = [ ( 1.838D × 2.76 ) + (8D × 1 ) ] ÷ [1.838D + 8D]

= 13.07288D ÷ 9.838D

= 1.329 m

or

= 132.9 cm