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Mass-Centriod-Moment

Probability Introduction :

Illustrations :

Q1) Electric charge is distributed over the disk  so that the charge density at (x,y) is  coulombs per sq.meters. Find total charge on the disk.

Solution:-

‘Q’ be the total charge.

Q = ∬ σ(x,y) dA
    =∬(13+x^2+y^2)dA

    =∫_0^2π∫_0^(√2)((13+ r^2 )r.dr.dθ

    = (13 + 1)(2π)

      = 28 coulombs

Q2) A lamina occupies the part of the disk  is the first quadrant and the density at each point is given by the function.

A)   What is the total mass ?

B)    What is the moment above x-axis?

C)    What is the moment above y-axis?

D)   Where is the centre of mass ?

Solution:-

        x^2+y^2≤16=4^2    0≤r≤4

                                            0≤θ≤π/2        (First Quadrant)

Density function    ρ(xy)= 3(x^2+y^2 )
                                              = 3r^2

                                        dA = r.dr.d

 

A)   Total mass (M)  =∬ρ(xy)dA

                            = ∫_0^(π/4)∫_0^4   (3r^2).r.dr.dθ

                            =∫_0^(π/4)∫_0^4   3r^3 dr.dθ

                             = 192(π/2)

                              = 96π

B)    M_x= ∬yρ(xy)dA

             =∫_0^(π/2)∫_0^4( r sin⁡θ).(3r^2)  r.dr.dθ

             = ∫_0^(π/2)∫_0^4 (sin⁡θ.3r^4 ). dr.dθ

             = (3/5)(1024)
              =3072/5

              

C)   M_y= ∬xρ(xy)dA

             =∫_0^(π/2)∫_0^4 (r cos⁡θ).(3r^2) r.dr.dθ

             = ∫_0^(π/2)∫_0^4  (cosθ.3r^4 )dr dθ

             = (3/5)(1024)
             =
3072/5

             

D)   x _c=Mx/M

          =(3072/5).(1/96π)

          =32/5π

  y _c= My/x

          =32/5π

Centre of mass(32/5π  ,32/5π)

 


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