CBSE Class 12-science Answered

What is the centre of mass of parabolic spandral ? Explain briefly with diagram

Asked by rushabhjain.a | 23 Oct, 2019, 07:16: PM

Expert Answer

Fig.(1) shows the parabolic spandrel , which is a triangular plate cut at the base with parabolic curve as shown in figure.

Let us consider that we are given the dimensions, length L and width b . Curvature parameter k of parabolic curve y² = k x is also known.

To get centre of mass (CM), let us consider area of spandrel is difference of area of triangle OAB and parabolic curve area ADB

By symmetry, CM is at axis of spandrel OD. Hence only we need to find the distance of CM from O.

Area of Triangle A_T = (1/2) L b ;

Parabolic curve area from its length CD and known curve parameter k is calculated as follows

Area of parabola =

begin mathsize 14px style integral subscript 0 superscript l 2 square root of k x end root d x space equals space 4 over 3 square root of k space l to the power of bevelled 3 over 2 end exponent end style

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

where l is length OC of parabolic curve area . In the given spandrel, we are given the length AC = b/2

OC and BC are related as AC²= k OC or ( b² / 4 ) = k l or l = b²/(4k) ...............(2)

using eqn.(2) , Area of parabola given in eqn.(1) is rewritten as ,

Area of parabola A_P= (4/3) √k l^3/2 = (4/3) √k [ b³ / (4k)^3/2 ] = (1/6) ( b³ / k ) ....................(3)

Let x be the distance of CM (refer fig.(2) ) from point O

Then by taking moment of mass with respect to O, we have,

A_S x + A_P d = A_T (2/3)L ..................(4)

where A_S , A_P and A_T are area of spandrel, parabolic area and Triangle respectively.

d is distance of CM of parabolic area from O. For triangle, we consider CM is at a distance (2/3)L from O

Area of parabolic curve region A_P is given in eqn.(3)

Area of spandrel A_S = A_T - A_P = (1/2) L b - (1/6) ( b³ / k )

Centre of mass of parabolic area is (3/5) of its length from its apex point

( Refer answer for your previous question , CM of parabolic lamina )

Hence d , which is the distance of CM of parabolic area from O

d = OC - CD - (3/5) CD = OC - (2/5) CD = { L - [ b² / (10k) ] }

Hence In eqn.(4), all values except x are known. Hence x cand be solved using eqn.(4)

I leave it to user to get the final answer by solving eqn.(4)

Answered by Thiyagarajan K | 24 Oct, 2019, 12:01: PM

CBSE Class 12-science Answered

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CBSE Class 12-science Answered

Concept Videos

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