Consider a semi-spherical bowl machined out of wood (a drawing of a semi-spherical bowl can be found in unit- 4 introduction, solved example 3). the outer radius of the bowl is 12.0 cm and it is uniformly 2.0cm thick. what is the total surface area of the bowl (in units of m2)?

The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)

Explanation:

The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.

a) Outer semi-sphere:

A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²

b) Inner semi-sphere:

A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²

c) Edge (Ring):

A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²

Therefore, the total surface area of the bowl is given by:

A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)

Changing units to m², as required in the problem, we get:

A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)

0.19 m/s and 0.45 m/s

because if the tool that you are using is not balanced it there would be false mass.