`y=x^(2/3), y=4` 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 `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 given by:

`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`

Given:`y=x^(2/3),y=4`

Refer to attached image. The plot of `y=x^(2/3)` is red in color.

The graphs intersect at the coordinates `(-8,4),(8,4)`

Let's evaluate the area of the bounded region,

`A=int_(-8)^8(4-x^(2/3))dx`

`A=2int_0^8(4-x^(2/3))dx`

`A=2[4x-x^(2/3+1)/(2/3+1)]_0^8`

`A=2[4x-3/5x^(5/3)]_0^8`

`A=2[4*8-3/5(8)^(5/3)]`

`A=2[32-3/5(2^3)^(5/3)]`

`A=2[32-3/5(2)^5]`

`A=2[32-96/5]`

`A=2[(160-96)/5]`

`A=(2(64))/5`

`A=128/5`

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

Since the graph is symmetrical about the y-axis,

So, `M_y=0` and `barx=0` ,

`M_x=rhoint_(-8)^8 1/2([4^2]-[x^(2/3)]^2)dx`

`M_x=rho/2int_(-8)^8(16-x^(4/3))dx`

`M_x=rho/2(2)int_0^8(16-x^(4/3))dx`

`M_x=rho[16x-(x^(4/3+1)/(4/3+1))]_0^8`

`M_x=rho[16x-3/7x^(7/3)]_0^8`

`M_x=rho[16*8-3/7(8)^(7/3)]`

`M_x=rho[128-3/7(2^7)]`

`M_x=rho[128-3/7(128)]`

`M_x=512/7rho`

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

`bary=(512/7rho)/(rho128/5)`

`bary=(512/7)(5/128)`

`bary=20/7`

The coordinates of the center of mass are `(0,20/7)`

 

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