`y=x^2, y=x^3` 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

For an irregularly shaped planar lamina of uniform density `(rho)` bounded by graphs `y=f(x),y=g(x)` and `a<=x<=b` , the mass `(m)` of this region is given by,

`m=rhoint_a^b[f(x)-g(x)]dx`

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

The moments about the x- and y-axes are,

`M_x=rhoint_a^b 1/2([f(x)]^2-[g(x)]^2)dx`

`M_y=rhoint_a^bx(f(x)-g(x))dx`

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

`barx=M_y/m`

`bary=M_x/m`

Now we are given `y=x^2,y=x^3`

Refer the attached image. Plot in red color is of `y=x^2` and blue color is of `y=x^3`

Curves intersect at (1,1)

Now let's evaluate the area of the region,

`A=int_0^1(x^2-x^3)dx`

`A=[x^3/3-x^4/4]_0^1`

`A=[1^3/3-1^4/4]`

`A=(1/3-1/4)=(4-3)/12`

`A=1/12`

Now let's evaluate the moments about the x- and y-axes,

`M_x=rhoint_0^1 1/2[(x^2)^2-(x^3)^2]dx`

`M_x=rho/2int_0^1(x^4-x^6)dx`

`M_x=rho/2[x^5/5-x^7/7]_0^1`

`M_x=rho/2[1^5/5-1^7/7]`

`M_x=rho/2(1/5-1/7)`

`M_x=rho/2(7-5)/(35)`

`M_x=rho/35`

`M_y=rhoint_0^1x(x^2-x^3)dx`

`M_y=rhoint_0^1(x^3-x^4)dx`

`M_y=rho[x^4/4-x^5/5]_0^1`

`M_y=rho[1^4/4-1^5/5]`

`M_y=rho[5-4)/(20)`

`M_y=rho/20`

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

Plug in the values of `M_y` and `A` ,

`barx=(rho/20)/(rho1/12)`

`barx=12/20`

`barx=3/5`

`bary=M_x/m=M_x/(rhoA)`

`bary=(rho/35)/(rho1/12)`

`bary=12/35`

The coordinates of the center of mass are `(3/5,12/35)`

 

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