Moment of Inertia of a Hemisphere

[hp-p00nst3r]

Homework Statement

The moment of inertia of a uniform sphere of mass m and radius r about an axis passing through its center of mass is (2/5) mr^2. If half of the sphere is removed, what is the moment of inertia about the same axis?

Homework Equations

integral of r^2 dm

The Attempt at a Solution

http://img407.imageshack.us/img407/5250/mech221ps4moiim8.jpg
I think this is wrong since I've looked up the actual I for this and it gives me (2/5) mr^2 still. I also tried using the parallel axis theorem on this and it also gives me the (2/5) mr^2. What am I doing wrong?

 

[alphysicist]

Hi hp-p00nst3r,

Homework Statement

The moment of inertia of a uniform sphere of mass m and radius r about an axis passing through its center of mass is (2/5) mr^2. If half of the sphere is removed, what is the moment of inertia about the same axis?

Homework Equations

integral of r^2 dm

The Attempt at a Solution

http://img407.imageshack.us/img407/5250/mech221ps4moiim8.jpg
I think this is wrong since I've looked up the actual I for this and it gives me (2/5) mr^2 still. I also tried using the parallel axis theorem on this and it also gives me the (2/5) mr^2. What am I doing wrong?

About the third thing you have written down is this:

$$dI = \frac{1}{2} z^2 dm$$

This should represent the moment of inertia of a representative disk, like the one you have in the diagram. However, the moment of inertia of a disk is (1/2) M R², where R is the radius of the disk.

From the figure, you can see that the radius of the disk is not z, so your expression for dI is not correct. The radius is actually y, so once you have used the expression for y in dI, I think you should get the right answer.

 

[baba89]

Your solution is incorrect because you are assuming that the moment of inertia of the hemisphere is the same as that of a full sphere. However, when half of the sphere is removed, the distribution of mass changes and the moment of inertia will also change.

To find the moment of inertia of the hemisphere with half of it removed, we can use the parallel axis theorem. The moment of inertia of the hemisphere about its own axis (passing through its center of mass) is (2/5)mr^2. We can then use the parallel axis theorem to find the moment of inertia about the same axis after removing half of the hemisphere.

Let's define the axis of rotation as the x-axis, with the origin at the center of mass of the hemisphere. The moment of inertia of the hemisphere about this axis is given by:

I = (2/5)mr^2

Now, let's consider the moment of inertia of the removed half hemisphere. We can calculate this by using the integral of r^2 dm, where r is the distance from the axis of rotation to the infinitesimal mass element dm. Since the removed half hemisphere has a constant density, we can write dm = ρdV, where ρ is the density and dV is the infinitesimal volume element. The moment of inertia of the removed half hemisphere can then be calculated as:

I' = ∫(x^2 + y^2)dm = ∫(x^2 + y^2)ρdV

Since the removed half hemisphere is symmetrical about the x-axis, we can rewrite this integral as:

I' = 2∫(x^2 + y^2)ρdV

Now, we can use the fact that the volume of the removed half hemisphere is half of the volume of the full hemisphere, and the density is constant, to write:

I' = 2(1/2)ρ∫(x^2 + y^2)dV = ρ∫(x^2 + y^2)dV

We can now use the parallel axis theorem to find the moment of inertia of the removed half hemisphere about the x-axis:

I' = ρ∫(x^2 + y^2)dV + mr^2 = ρ∫(x^2 + y^2)dV + (1/2)mr^2

Finally, we can substitute this expression for I'