Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

9321924448 / 9987178554

Mon to Sat - 10 AM to 7 PM

What is the centre of mass of sector of the circle. Explain with diagram

Asked by rushabh1234 17th October 2019, 6:05 AM
Answered by Expert
Answer:
First let us find the centre of mass of thin rod bent in the form of circular arc that subtends angle 2θ at centre.
 
Let us cinsider a small element of length dl that subtends angle dθ at centre.
Let mass is distributed uniformly and λ is mass per unit length of rod .
 
mass moment of this small element along x-axis :-  ( R dθ λ ) ( R cosθ) = R2 λ cosθ dθ
 
mass moment of this small element along y-axis :-  ( R dθ λ ) ( R sinθ) = R2 λ sinθ dθ
 
x-coordinate of centre of mass of ring :- begin mathsize 14px style fraction numerator R squared lambda over denominator M end fraction integral subscript negative theta end subscript superscript theta cos theta apostrophe space d theta apostrophe space equals space 2 fraction numerator R squared lambda over denominator M end fraction sin theta end style ........................... (1)
y-coordinate of centre of mass of ring :- begin mathsize 14px style fraction numerator R squared lambda over denominator M end fraction integral subscript negative theta end subscript superscript theta sin theta apostrophe space d theta apostrophe space equals space 0 end style   ..............................(2)
Where M is mass of whole rod.
 
Mass density λ is defined as, λ =  mass / length = M/( R 2θ )  ..................(3)
 
using the expression for λ in (3), we get x-coordinate and y-coordinate of centre of mass from eqn.(1) and (2) as
 
Centre of mass :-  [ ( R sinθ / θ ) , 0 ] .................................(4)
 
Coordinates are given in above equation, as shown in figure, considering centre of curvature of arc coincides with origin and
line connecting the origin and centre of circular arc coincides with x-axis.
-------------------------------------------------------------------
We consider sector of circular disc , angle 2θ at centre , is made up sevaral rings in the shpae of circular arc as shown in figure.
 
Since we calculated coordinates of centre of mass of circular arc, using that formula  for a circular arc ring of thickness dr,
we find the x-coordinate of centre of mass of sector of circular disc as follows
 
mass of circular arc ring = r×2θ×dr×ρ
 
where ρ is density per unit area
 
If M is total mass of sector of circular disc, then x-coordinate of centre of mass of sector of circular disc is given as
 
begin mathsize 14px style 1 over M integral subscript 0 superscript R r space cross times 2 theta cross times d r cross times rho cross times r cross times fraction numerator sin theta over denominator theta end fraction space equals space fraction numerator 2 space sin theta space rho over denominator M end fraction integral subscript 0 superscript R r squared d r space equals space 2 over 3 fraction numerator rho space R cubed over denominator M end fraction sin theta end style...............(5)
density ρ is defined as, ρ = M/ [ (2θ/2π) πR2 ] = M / [ θ R2 ] ...............(6)
 
By substituting ρ in eqn.(5),  x-coordinate of centre of mass of sector of circular disc is given as   (2/3)R (sinθ/θ)
 
x and y coordinate of centre of mass of sector of circular disc is given as  [ (2/3)R (sinθ/θ) , 0 ]
 
Coordinates are given in above equation, as shown in figure, considering centre of curvature of sector of circle coincides with origin and
midline dividing the sector of circle coincides with x-axis.
 
Answered by Expert 17th October 2019, 10:23 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp