You'd see that it doesn't come to $\frac MR^2$. (You can verify this by trying to calculate the volume of the sphere using your formula. So, the shape that you have described is not a disc at all. Furthermore, because of the symmetry of the sphere, each principal moment is the same, so the moment of inertia of the sphere taken about any diameter is. Moment of Inertia of hollow sphere Trouble calculating the moment of inertia of a solid cylinder with axis of rotation passing perpendicularly through the. The parallel axis theorem can also be used to find a centroidal moment of inertia when you already know the moment of inertia of a shape about another axis, by using the theorem ‘backwards’, I I + Ad3 I I Ad2. (1) the moment of inertia tensor is (2) (3) (4) which is diagonal, and so it is in principal axis form. The result is clearly different, and shows you cannot just consider the mass of an object to be concentrated in one point (like you did when you averaged the distance). The dimensions of the ring are Ri 30 mm, Ro 45 mm, and a 80 mm. $Rd\theta$ is the length of a tiny, tiny arc of radius $R$ and angle $d\theta$ and in the infinitesimal limit it can be approximated to a straight line, that is, a chord but notice that this chord is still not along the z axis (id est, the vertical axis). The total moment of inertia is just their sum (as we could see in the video): I i1 + i2 + i3 0 + mL2/4 + mL2 5mL2/4 5ML2/12. That is, a body with high moment of inertia resists angular acceleration, so if it is not. The moment of inertia expresses how hard it is to produce an angular acceleration of the body about this axis. m mass of sphere (kg, slugs) r radius in sphere (m, ft) Rectangular Plane. The moment of inertia of a body, written IP, a, is measured about a rotation axis through point P in direction a. However, the thickness of this differential disc is NOT $ R d\theta$ but $Rd\theta cos\theta$. A generic expression of the inertia equation is. The differential mass element in this case is a disc, of radius $r$ where $r = R \cos\theta$ as you have correctly used. The mistake is in the second line, in the calculation of the differential mass element. If the moment of inertia of a rigid body about an axis through its center of mass is given by I c m, then the moment of inertia around an axis parallel to the original axis and separated from it by a distance d is given by (5.4.5) I I c m + m d 2 where m is the object’s mass. For a uniform disk of radius r and total mass m the moment of inertia is simply 1/2 m r2.
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