# Moment of inertia of 1/4 of a sphere [closed]

Is it even possible to calculate moment of inertia of strange objects like 1/4 of a sphere?

• Moment of inertia around which axis? Commented Jan 7, 2021 at 3:16
• Moment of inertia along the z-axis and at the geometric center (and not the center of mass)? Commented Jan 7, 2021 at 5:58

Yes. It is always possible to calculate the moment of inertia of any arbitrary distribution of mass by doing a volume integral, summing up the moment of inertia for every infinitesimal mass $$dM=\rho\,dV$$:

$$I=\int dM\,r^2 =\int \rho\,dV\,r^2.$$

Here $$r$$ is the distance of each infinitesimal mass from the axis of rotation.

Sometimes this integral cannot be done analytically and must be done numerically.

Sometimes there are symmetry arguments that allow one to calculate the moment of inertia for one shape from that for a simpler shape, without having to integrate over a complicated geometry.

Yes. If you know the shape and density you can calculate the inertia about a fixed axis or about a point. I have edited this answer to respond to comments received.

To evaluate rotation about a fixed axis (or more generally, general rotation in a plane), you use the moment of inertia about that axis; this moment of inertia is a scalar. Let the axis of rotation be the $$z$$ axis; $$J_z = I_z\omega$$ where $$J_z$$ is the $$z$$ component of the angular momentum, I is the moment of inertia with respect to the $$z$$ axis, and $$\omega$$ is the angular velocity that is along the $$z$$ axis. We do not provide specific answers to specific homework-type questions on this site. A good physics text, such as one by Halliday and Resnick, provides the necessary discussion and relationships, including how to evaluate the moment of inertia, that you can apply to any object such as 1/4 of a sphere.

(As pointed out in a comment by @Michael Seifert, if the fixed axis of rotation is not a principal axis, the angular momentum may have components along the other- $$x$$ and $$y$$ axes- as well; but $$J_z = I_z\omega$$ is still valid .)

To evaluate general three-dimensional rotation about a point, you use the inertia tensor: $$\vec J = \buildrel\leftrightarrow\over I\cdot\vec\omega$$, where $$\buildrel\leftrightarrow\over I$$ is the inertia tensor, $$\vec J$$ is the angular momentum vector, and $$\vec\omega$$ is the angular velocity vector.) If the rotation is constrained to be about a fixed point, that point is taken as the origin of the coordinate system for evaluating the rotational motion. If the object is unconstrained and moving freely, the center of mass (even if accelerating) is taken as the origin of the coordinate system for evaluating the rotational motion; translational motion is evaluated for the center of mass. If the body is constrained to rotate about a fixed point and that point is accelerating in an inertial frame, the evaluation must consider the fictitious forces present using that point as the origin. A good intermediate/advanced physics mechanics book addresses the general motion of a rigid body; for example, see Mechanic by Symon, or Classical Mechanics, by Goldstein.

For a complicated shape the calculation may have to be done numerically.

• Nitpick: if you rotate an object about a fixed but non-principal axis, you generally can't treat the moment of inertia as a scalar, since you will not have $\vec{L}$ parallel to $\vec{\omega}$. Commented Jan 6, 2021 at 17:51
• @Michael Seifert thanks for the comment. I modified my answer accordingly. Commented Jan 6, 2021 at 20:32

Yes, use symmetry arguments for this particular case(think of how this setup is different from a sphere whose $$I_{CM}$$ is known). Otherwise in general case integratation is always an option, it may be complicated or you may need to solve the integral numerically but it will definetely give you the answer.

Of course you can! You "simply" have to use the definition of moment of inertia. The difficult is doing the right integration. Remember to specificate always respect which axis are you calculating the moment of inertia to. In the picture it's shown. Also, should be specified if the rigid body has an uniform density. Let assume it is. In this case you should use the Huygens-Steiner Theorem: first calculate the moment of inertia respect the axis that crosses the CM (that is the smallest moment of inertia you can calculate of every rigid body, and it's easier to evaluate) and then use the theorem. Sometimes symmetries can be a salvation for the evaluation of the integral. Hope this can be helpful to give you the principal idea.

If you look carefully at the Integral you want to perform, you notice there is no angle dependence. So you can move pieces of you sphere around, as long as r of each individual element stays the same. By using rotations, you can rotate each of the 'quarter' elements of the sphere to exactly match your object. Long story short this means: I(Sphere)=4I(your shape)