Moments of Inertia and More....2

[harpazo]

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

The diagram given is a circle with radius a. The entire circle is shaded.

We are also given I_o = (pi•a^4)/2, where I_o is the moment of inertia about the origin.

My Work:

I found the mass to be 4pi.

I found I_y to be 8pi.

x double bar = sqrt{8pi/4pi}

x double bar = sqrt{8/4}

x double bar = sqrt{2}

I found I_x = 16pi.

y double bar = sqrt{16pi/4pi}

y double bar = sqrt{4}

y double bar = 2.

Is any of this correct?

 

[MarkFL]

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

The diagram given is a circle with radius a. The entire circle is shaded.

We are also given I_o = (pi•a^4)/2, where I_o is the moment of inertia about the origin.

So, the coordinate axes are oriented such that the origin is at the center of the disk. The moment of inertia where the axis of rotation is the $z$-axis is given by:

$$I_z=\frac{mr^2}{2}$$

The mass $m$ for this disk is:

$$m=\pi a^2$$

And the radius $r$ is stated to be $a$.

Hence:

$$I_z=\frac{\pi a^2\cdot a^2}{2}=\frac{\pi a^4}{2}$$

My Work:

I found the mass to be 4pi.

The mass should be in terms of the parameter $a$, since the density is constant. I gave the mass above.

I found I_y to be 8pi.

You should be able to show that:

$$I_x=I_y=\frac{\pi a^4}{4}$$

 

[harpazo]

So, the coordinate axes are oriented such that the origin is at the center of the disk. The moment of inertia where the axis of rotation is the $z$-axis is given by:

$$I_z=\frac{mr^2}{2}$$

The mass $m$ for this disk is:

$$m=\pi a^2$$

And the radius $r$ is stated to be $a$.

Hence:

$$I_z=\frac{\pi a^2\cdot a^2}{2}=\frac{\pi a^4}{2}$$
The mass should be in terms of the parameter $a$, since the density is constant. I gave the mass above.
You should be able to show that:

$$I_x=I_y=\frac{\pi a^4}{4}$$

Interestingly tricky.