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CBSE Class 11-science Answered

What is the centre of mass of arc of semicircle and arc of circle?explain briefly with diagram

Asked by rushabhjain.a | 24 Oct, 2019, 01:52: PM
Expert Answer

Figure-1 shows a semi circle of radius R. Let the centre of curvature of arc coincide with the origin of the coordinate system.
 
Let us consider a small element of length dl at angular position θ and this elment of length dl subtends angle dθ.
 
Lenth of the small element = dl = R dθ .
 
Let ρ be the linear density , i.e. mass per unit length. Mass of the small element = R dθ × ρ
 
By symmetry, centre of mass (CM) is along y-axis, i.e, x-coordinate  CX of CM is zero.
 
Y-coordinate CY of Centre of mass of semi-circular arc is obtained as
 
begin mathsize 14px style C subscript Y space equals space 1 over M integral subscript 0 superscript pi R space d theta space rho space R space sin theta space space equals space fraction numerator R squared rho over denominator M end fraction integral subscript 0 superscript pi sin theta space d theta space equals space fraction numerator 2 R squared rho over denominator M end fraction end style ................(1)
where M is total mass of semi-circular arc .  we have,  M = ρ×π R
 
By substituting for mass M in eqn.(1), we get,  CY = 2R/π  .......................(2)
 
Hence Centre of mass of semi-circle as shown in fig.(1) :- [ CX , CY ] = [ 0, (2R)/π ]
 
Centre of mass is at symmetry axis at a distance (2R)/π from centre of curvature of semi-circle.
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Figure-2 shows a circular arc that subtends angle φ at centre . 
Let the axis of symmetry coincide with x-axis of coordinate system and centre of curvature of circular arc coincide with origin.
 
Let us consider small element of arc of legth dl at angular position θ.
Let this small element subtends angle dθ at centre as shown in figure.
 
By symmetry, centre of mass is at x-axis, i.e., y-coordinate CY of centre of mass is zero.
x-coordinate Cx is obtained as follows
 
begin mathsize 14px style C subscript X space equals space 1 over M integral subscript negative phi divided by 2 end subscript superscript phi divided by 2 end superscript R d theta space rho space R space cos theta space equals space fraction numerator R squared rho over denominator M end fraction integral subscript negative phi divided by 2 end subscript superscript phi divided by 2 end superscript cos theta d theta space equals space fraction numerator R squared rho over denominator M end fraction 2 space sin phi over 2 end style .......................(1)
In the above eqn. mass of arc ,  M = R φ ρ . By substituting mass in eqn. (1), we get,  CX = ( 2R/φ ) sin(φ/2) 
 
Hence centre of mass of arc as shown in figure-2 is,  [ CX , CY ] = [ ( 2R/φ ) sin(φ/2) , 0 ]
Answered by Thiyagarajan K | 24 Oct, 2019, 09:42: PM
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