The radius of gyration K of a hollow sphere o | vedclass

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

Question

(A) The moment of inertia of a hollow sphere about an axis passing through its center is $I_{cm} = \frac{2}{3}MR^2$.By the parallel axis theorem, the

Vedclass · Accepted Answer

$R/\sqrt{3}$

Vedclass · Answer

(A) The moment of inertia of a hollow sphere about an axis passing through its center is $I_{cm} = \frac{2}{3}MR^2$.By the parallel axis theorem, the moment of inertia $I$ about an axis at a distance $h$ from the center is given by $I = I_{cm} + Mh^2$.Substituting the value of $I_{cm}$, we get $I = \frac{2}{3}MR^2 + Mh^2$.The radius of gyration $K$ is defined by the relation $I = MK^2$.Given that $K = R$, we have $I = MR^2$.Equating the two expressions for $I$:$MR^2 = \frac{2}{3}MR^2 + Mh^2$$R^2 = \frac{2}{3}R^2 + h^2$$h^2 = R^2 - \frac{2}{3}R^2$$h^2 = \frac{1}{3}R^2$$h = \frac{R}{\sqrt{3}}$Thus, the correct option is $A$.

The radius of gyration $K$ of a hollow sphere of mass $M$ and radius $R$ about an axis $XY$ is equal to $R$. The distance of that axis from the center of the sphere is $h$. The value of $h$ is

Similar Questions

Each spherical shell has a radius $R$ and mass $M$. They are connected by a light massless rod. Calculate the moment of inertia about the axis $xx'$.

The moment of inertia of a disc about its axis is $I$. What will be its moment of inertia about a tangent in its plane?

The moment of inertia of a uniform circular disc about its diameter is $I$. What is its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim (in $, I$)?

Consider a uniform horizontal solid cylinder of mass $10 \,kg$ such that its length is $9$ times its radius. Let the radius be $40 \,cm$. Calculate the moment of inertia of the cylinder about a line passing through its edge and perpendicular to its axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module