Moment of inertia of a sphere (Is this right?)

[een]

Homework Statement

A homogenous punctured sphere with inner radius a and outer radius b rotates around an axis that touches its outer surface. Determine the moment of inertia. [hint: use the parallel axis theorem]

Homework Equations

The Attempt at a Solution

basically i used... {($$\rho$$,$$\theta$$), a<$$\rho$$<b, 0<$$\theta$$<2pi), d(p,c) =$$\rho$$cos$$\theta$$

I = $$\int$$0-2pi$$\int$$a-b (M/(pi(b^2-a^2)))$$\rho$$^2*cos^2$$\theta$$*$$\rho$$*d$$\rho$$d$$\theta$$

I simplified and got I = M(5b^2+a^2)/4...IS THIS RIGHT?

 

[mighty2000]

Thank you for your question. Your attempt at solving the problem is on the right track, but there are a few errors in your calculation.

Firstly, the parallel axis theorem states that the moment of inertia of a body about an axis parallel to its center of mass is equal to the moment of inertia about the center of mass plus the product of the mass and the square of the distance between the two axes. In this case, the distance between the axis of rotation and the center of mass is the outer radius b. Therefore, the correct formula for the moment of inertia is:

I = Icm + Mb^2

Where Icm is the moment of inertia about the center of mass, given by:

Icm = (2/5)Ma^2

Substituting this into the first equation, we get:

I = (2/5)Ma^2 + Mb^2

Simplifying, we get:

I = (2/5)M(a^2 + 5b^2)

Therefore, the correct answer for the moment of inertia is:

I = (2/5)M(a^2 + 5b^2)

I hope this helps clarify any confusion and good luck with your further studies in physics. Keep up the good work!