What is the moment of inertia of a disc about | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is $I_z = \frac{MR^2}{2}$,where $M$ is the

Vedclass · Accepted Answer

$ \frac{MR^2}{4} $

Vedclass · Answer

(B) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is $I_z = \frac{MR^2}{2}$,where $M$ is the mass and $R$ is the radius of the disc.Since the disc is a planar body,we can apply the theorem of perpendicular axes.Let the $x$ and $y$ axes lie in the plane of the disc,intersecting at the centre $O$. These axes represent two diameters of the disc.According to the theorem of perpendicular axes,$I_z = I_x + I_y$.Due to the rotational symmetry of the disc,the moment of inertia about any diameter is the same,so $I_x = I_y = I_d$.Substituting this into the theorem,we get $I_z = I_d + I_d = 2I_d$.Therefore,$I_d = \frac{I_z}{2} = \frac{1}{2} \left( \frac{MR^2}{2} \right) = \frac{MR^2}{4}$.Thus,the moment of inertia of a disc about one of its diameters is $\frac{MR^2}{4}$.

What is the moment of inertia of a disc about one of its diameters?

Similar Questions

If an object lies in the $X-Y$ plane,then according to the Perpendicular Axis Theorem:

From a circular disc of radius $R$ and mass $9M$,a small disc of radius $\frac{R}{3}$ is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is:

$A$ rigid body of mass $m$ is rotating with angular velocity $\omega$ about an axis at a distance '$a$' from the center of mass $G$. The radius of gyration about an axis passing through $G$ is $K$. The rotational kinetic energy of this body about the new parallel axis will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module