`x=4-y^2 , x=0` Find the x and y moments of inertia and center of mass for the laminas of uniform density `p` bounded by the graphs of the equations.

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gsarora17 eNotes educator| Certified Educator

Consider an irregularly shaped planar lamina of uniform density `rho` , bounded by graphs `x=f(y)` , `x=g(y)` and `c<=y<=d` . The mass `m` of this region is given by:

`m=rhoint_c^d(f(y)-g(y))dy`

`m=rhoA` , where A is the area of the region

The moments about the x- and y-axes are given by:

`M_x=rhoint_c^d y(f(y)-g(y))dy`

`M_y=rhoint_c^d 1/2([f(y)]^2-[g(y)]^2)dy`

The center of mass `(barx,bary)` is given by:

`barx=M_y/m`

`bary=M_x/m`

We are given `x=4-y^2` ,`x=0`  

Refer to the attached image. The plot of `x=4-y^2` is red in color.

Let's evaluate the area of the region,

`A=int_(-2)^2(4-y^2)dy`

`A=2int_0^2(4-y^2)dy`

`A=2[4y-y^3/3]_0^2`

`A=2[4*2-2^3/3]`

`A=2[8-8/3]`

`A=32/3`

`M_y=2rhoint_0^2 1/2(4-y^2)^2dy`

`M_y=(2rho)/2int_0^2(4^2-2(4)y^2+(y^2)^2)dy`

`M_y=rhoint_0^2(16-8y^2+y^4)dy`

`M_y=rho[16y-8(y^3/3)+y^5/5]_0^2`

`M_y=rho[16*2-8/3(2^3)+1/5(2^5)]`

`M_y=rho[32-64/3+32/5]`

`M_y=rho[(480-320+96)/15]`

`M_y=256/15rho`

By symmetry, `M_x=0,bary=0`

`barx=M_y/m=M_y/(rhoA)`

`barx=(256/15rho)/(rho32/3)`

`barx=(256/15)(3/32)`

`barx=8/5`

The coordinates of the center of mass are `(barx,bary)` are `(8/5,0)`  

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