Find the moment of inertia of the ring shown | vedclass

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

Question

(A) The moment of inertia of a ring about its diameter is given by $I_{dia} = \frac{1}{2} MR^2$.The axis $AB$ is parallel to the diameter of the ring

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The moment of inertia of a ring about its diameter is given by $I_{dia} = \frac{1}{2} MR^2$.The axis $AB$ is parallel to the diameter of the ring and is at a distance $R$ from the center of the ring.According to the parallel axis theorem,the moment of inertia about an axis parallel to the diameter is $I_{AB} = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about the center of mass (which is the diameter in this case) and $d = R$ is the distance between the axes.Substituting the values,we get:$I_{AB} = \frac{1}{2} MR^2 + MR^2$$I_{AB} = \frac{3}{2} MR^2$

Find the moment of inertia of the ring shown in the figure about the axis $AB$.

Similar Questions

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

Moment of inertia of a disc about its own axis is $I$. Its moment of inertia about a tangential axis in its plane is

$A$ square plate of mass $m$ and side $a$ has a moment of inertia about an axis perpendicular to its plane and passing through one of its corners equal to $\alpha \,ma^2$. Then $\alpha$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module