CBSE Class 12-science Answered

Find the Center of gravity

Asked by sahilsah9211 | 21 Jun, 2023, 01:58: PM

Expert Answer

Centre of gravities of semi-circular segment, circular segment and triangle are show in the following figure.

For a semi circular segment , centre of gravity is at symmetry axis and at a distance (2/3)R from diameter , where R is radius of semi-circle.

For a circular segment , centre of gravity is at symmetry axis and at a distance (2/3)R from centre of curvature , where R is radius of semi-circle.

For a triangle , centre of gravity is at the median of triangle and at a distance (1/3)h from base .

The given shape can be divided into six areas each having a standard shape like semi-circle, circular segment and triangle as shown below

Area A₁ is semi-circle of radius 4m . Area A₂ and A₃ are circular segment of radius 4 m .

Area A₄ is isoceless triangle of height 3 m and base 6m. Area A₅ is isoceless triangle of height 4 m and base 6 m.

Area A₆ is cut-out portion from the given shape which is a semicircle of radius 2m.

As shown in figure , given shape is symmetrically placed on y-axis . Hence x-coordinate of centre of gravity is zero.

Y-coordinate of center of gravity Y is calculated from the following equation.

begin mathsize 14px style A subscript T space Y space equals space A subscript 1 y subscript 1 plus A subscript 2 y subscript 2 plus A subscript 3 y subscript 3 plus A subscript 4 y subscript 4 plus A subscript 5 y subscript 5 minus A subscript 6 y subscript 6 space end style

.........................(1)

Where A₁ , A₂ , .......A₅ and A₆ are respective areas of parts and y₁, y₂, ............y₅ and y₆ are

y-coordinates of centre of gravity of respective part.

A_T is total area.

begin mathsize 12px style A subscript 1 space equals space 1 half pi cross times 4 squared space equals space 8 straight pi space space m squared space space space semicolon space space space straight y subscript 1 space equals space 4 divided by 3 space straight m space end style

begin mathsize 14px style A subscript 2 space equals space 1 over 8 straight pi cross times 4 squared space equals space 2 straight pi space straight m squared space space semicolon space space straight y subscript 2 space equals space 4 space plus space 2 over 3 cross times 4 space cross times sin left parenthesis 22.5 right parenthesis space equals space 5.02 space m end style

begin mathsize 14px style A subscript 3 space equals space 1 over 8 straight pi cross times 4 squared space equals space 2 straight pi space straight m squared space space semicolon space space straight y subscript 3 space equals space 4 space plus space 2 over 3 cross times 4 space cross times sin left parenthesis 22.5 right parenthesis space equals space 5.02 space m end style

begin mathsize 14px style A subscript 4 space equals space 1 half cross times 6 cross times 3 space equals space 9 space m squared space semicolon space space y subscript 4 space equals space 4 space plus space 2 over 3 cross times 3 space equals space 6 space m end style

begin mathsize 12px style A subscript 5 space equals space 1 half cross times 6 cross times 4 space equals space 12 space m squared space semicolon space space space space y subscript 5 space equals space 7 space plus space 1 third cross times 4 space space equals space 8.33 space m end style

begin mathsize 14px style A subscript 6 space equals space 1 half straight pi cross times 2 squared space equals space 2 straight pi space straight m squared space space space semicolon space space space straight y subscript 6 space equals space 1 third cross times 2 space equals space 0.67 space straight m end style

Total area A_Tis

begin mathsize 14px style A subscript T space equals space A subscript 1 plus A subscript 2 plus A subscript 3 plus A subscript 4 plus A subscript 5 minus A subscript 6 space equals space 52.42 space m squared end style

By substituting relevant values in eqn.(1), we get y-coordinate of centre of gravity as

begin mathsize 14px style Y space equals space 4.7 space m end style

Answered by Thiyagarajan K | 23 Jun, 2023, 12:22: AM

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