MHB Moments of Inertia and More....1

[harpazo]

Verify the given moment(s) of inertia and find x double bar and y double bar. Assume that the given lamina has a density of p = 1, where p is rho.

I_x = (bh^3)/3

I_y = (b^3h)/3

I found the mass to be bh.

x double bar = sqrt{(bh^3)/3 ÷ bh}

x double bar = sqrt{b^2/3}

x double bar = [(b•sqrt{3})/3]

y double bar = sqrt{(b^3h)/3 ÷ bh}

y double bar = sqrt{h^2/3}

y double bar = [(h•sqrt{3}/3]

Note: The diagram given for this problem is a rectangle in quadrant 1 from 0 to b along the x-axis and 0 to h along the y-axis.

Is any of this correct?

 

[MarkFL]

We have the definition:

$$I_x=\iint\limits_{R}y^2\rho(x,y)\,dA$$

Now, we are told:

$$\rho(x,y)=1$$

And so we have:

$$I_x=\iint\limits_{R}y^2\,dA$$

Using the given rectangular region of the lamina, this becomes:

$$I_x=\int_0^b\int_0^h y^2\,dy\,dx=\frac{h^3}{3}\int_0^b \,dx=\frac{bh^3}{3}$$

Using the definition:

$$I_y=\iint\limits_{R}x^2\rho(x,y)\,dA$$

Can you now find $I_y$ for the given lamina?

 

[harpazo]

We have the definition:

$$I_x=\iint\limits_{R}y^2\rho(x,y)\,dA$$

Now, we are told:

$$\rho(x,y)=1$$

And so we have:

$$I_x=\iint\limits_{R}y^2\,dA$$

Using the given rectangular region of the lamina, this becomes:

$$I_x=\int_0^b\int_0^h y^2\,dy\,dx=\frac{h^3}{3}\int_0^b \,dx=\frac{bh^3}{3}$$

Using the definition:

$$I_y=\iint\limits_{R}x^2\rho(x,y)\,dA$$

Can you now find $I_y$ for the given lamina?

I can find I_y. I will post my work later tonight.