Student Question

`y= 1/2x^2 , y=0, x=2` 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.

Expert Answers

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The center of Mass is:

`(x_(cm),y_(cm))=(M_y/M, M_x/M)`

Where the moments of mass are defined as:

`M_x=int int_A rho(x,y)*y dy dx`

`M_y=int int_A rho(x,y)*x dy dx`

The total mass is defined as:

`M=int int_A rho(x,y)dy dx`

First, lets find the total mass.

`M=int^2_0 [int^(1/2x^2)_0 rho dy] dx`

`M=rho int^2_0 [y|^(1/2x^2)_0] dx`

`M=rho int^2_0 1/2x^2 dx`

`M=rho/2 (1/3)x^3|^2_0`

`M=rho/2 (1/3)2^3`

`M=4/3 rho`

Now lets find the x moment of mass.

`M_x=int^2_0 [int^(1/2x^2)_0 rho*y dy] dx`

`M_x=rho int^2_0 [(1/2)y^2|^(1/2x^2)_0] dx`

`M_x=rho int^2_0 [(1/2)(1/2x^2)^2] dx`

`M_x=rho int^2_0 (1/8)x^4 dx`


`M_x=rho/40*2^5=32/40 rho=4/5 rho`

Now the y moment of mass.

`M_y=int^2_0 [int^(1/2x^2)_0 rho*x dy] dx`

`M_y=rho int^2_0 x[int^(1/2x^2)_0 dy] dx`

`M_y=rho int^2_0 x[y|^(1/2x^2)_0] dx`

`M_y=rho int^2_0 x[1/2x^2] dx`

`M_y=rho/2 int^2_0 x^3 dx`

`M_y=rho/2 (1/4)x^4|^2_0`

`M_y=rho/2 (1/4)2^4=2rho`

Therefore the center of mass is:

`(x_(cm),y_(cm))=(M_y/M, M_x/M)=((2rho)/(4/3 rho),(4/5 rho)/(4/3 rho))=(3/2,3/5)`

The moments of inerita or the second moments of the lamina are:

`I_x=int int_A rho(x,y)*y^2 dy dx`

`I_y=int int_A rho(x,y)*x^2 dy dx`

I won't solve these integrals step by step since they are very similar to the others, but you will find that:

`I_x=16/21 rho`

`I_y=16/5 rho`

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