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What is the centre of mass of pyramid? Explain briefly with examples
Asked by polyrelation | 21 Oct, 2019, 22:11: PM
answered-by-expert Expert Answer
A regular pyramid has square base and triangular faces as lateral sides. 
Let us have x-y-z coordinate system with origin coincide with centre of square base.
let x-y plane is parallel to square base and z-axis coincides with vertical axis of pyramid.
 
centre of mass of 3-dimensional solid is given as
 
begin mathsize 14px style z subscript c m space end subscript equals space fraction numerator integral subscript z m i n end subscript superscript z m a x end superscript integral subscript y m i n end subscript superscript y m a x end superscript integral subscript x m i n end subscript superscript x m a x end superscript z d x d y d z over denominator integral subscript z m i n end subscript superscript z m a x end superscript integral subscript y m i n end subscript superscript y m a x end superscript integral subscript x m i n end subscript superscript x m a x end superscript d x d y d z end fraction end style
where zmin and zmax are lower and upper limits of integration. For our pyramid, zmin = 0, zmax = h.
 
ymax and ymin are y-limits.   xmax and xmin are x-limits.
 
x and y limits are obtained from the sides of square cross section at a given height as shown below
 
 
 
centre of mass of pyramid is obtained  as begin mathsize 14px style z subscript c m space end subscript equals space fraction numerator integral subscript 0 superscript h integral subscript negative left square bracket 1 minus left parenthesis z divided by h right parenthesis right square bracket left parenthesis a divided by 2 right parenthesis end subscript superscript left square bracket 1 minus left parenthesis z divided by h right parenthesis right square bracket left parenthesis a divided by 2 right parenthesis end superscript integral subscript negative left square bracket 1 minus left parenthesis z divided by h right parenthesis right square bracket left parenthesis a divided by 2 right parenthesis end subscript superscript left square bracket 1 minus left parenthesis z divided by h right parenthesis right square bracket left parenthesis a divided by 2 right parenthesis end superscript z d x d y d z over denominator integral subscript 0 superscript h integral subscript negative left square bracket 1 minus left parenthesis z divided by h right parenthesis right square bracket left parenthesis a divided by 2 right parenthesis end subscript superscript left square bracket 1 minus left parenthesis z divided by h right parenthesis right square bracket left parenthesis a divided by 2 right parenthesis end superscript integral subscript negative left square bracket 1 minus left parenthesis z divided by h right parenthesis right square bracket left parenthesis a divided by 2 right parenthesis end subscript superscript left square bracket 1 minus left parenthesis z divided by h right parenthesis right square bracket left parenthesis a divided by 2 right parenthesis end superscript d x d y d z end fraction end style
from the above integration, we get zCM = h/4
 
Hence centre of mass of regular pyramid is at axis and at a height h/4 from base, where h is height of pyramid
Answered by Thiyagarajan K | 22 Oct, 2019, 14:31: PM

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