The moment of inertia of a uniform ring of ma | vedclass

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

Question

(B) The moment of inertia of a uniform ring about an axis passing through its center and perpendicular to its plane is $I_{cm} = Mr^2$.According to th

Vedclass · Accepted Answer

$\frac{3}{2}Mr^2$

Vedclass · Answer

(B) The moment of inertia of a uniform ring about an axis passing through its center and perpendicular to its plane is $I_{cm} = Mr^2$.According to the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the parallel axes.For a tangent in the plane of the ring,the distance $d$ between the center of mass axis and the tangent axis is equal to the radius $r$.Substituting the values,we get $I = Mr^2 + M(r)^2 = Mr^2 + Mr^2 = 2Mr^2$.Wait,the question asks for a tangent in its own plane. The moment of inertia about a diameter is $I_{diameter} = \frac{1}{2}Mr^2$.Using the parallel axis theorem for the tangent in the plane: $I = I_{diameter} + Md^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$.

The moment of inertia of a uniform ring of mass $M$ and radius $r$ about a tangent in its own plane is:

Similar Questions

Consider a thin uniform square sheet made of a rigid material. If its side is $a$,mass is $m$,and the moment of inertia about one of its diagonals is $I$,then:

$A$ sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is

Three rings each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about the axis $YY'$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module