The ratio of the radius of gyration of a holl | vedclass

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

Question

(C) For a hollow sphere of mass $M$ and radius $R$,the moment of inertia about its diameter axis $AB$ is $I_{	ext{sphere}} = \frac{2}{3} MR^2$.Since

Vedclass · Accepted Answer

$67$

Vedclass · Answer

(C) For a hollow sphere of mass $M$ and radius $R$,the moment of inertia about its diameter axis $AB$ is $I_{	ext{sphere}} = \frac{2}{3} MR^2$.Since $I = Mk^2$,the radius of gyration $k_1$ is given by $k_1^2 = \frac{2}{3} R^2$.For a solid cylinder of mass $M$,radius $R_{cyl} = R$,and length $L = 4R$,the moment of inertia about the axis $AB$ passing through the edge is calculated using the parallel axis theorem.The moment of inertia about the central longitudinal axis is $I_{cm} = \frac{1}{2} MR^2$.The distance from the central axis to the edge axis $AB$ is $d = R$.Using the parallel axis theorem,$I_{	ext{cylinder}} = I_{cm} + Md^2 = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.Wait,re-evaluating the cylinder: The axis $AB$ is at the edge of the circular face. The moment of inertia of a solid cylinder about its central longitudinal axis is $I_{cm} = \frac{1}{2} MR^2$. The distance from the center of mass to the axis $AB$ is $R$. Thus,$I_{AB} = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.However,based on the provided solution logic: $I_{	ext{cylinder}} = \frac{1}{12} M(4R)^2 + \frac{1}{4} MR^2 + M(2R)^2 = \frac{16}{12} MR^2 + \frac{1}{4} MR^2 + 4MR^2 = (\frac{4}{3} + \frac{1}{4} + 4) MR^2 = (\frac{16+3+48}{12}) MR^2 = \frac{67}{12} MR^2$.This corresponds to a cylinder rotating about an axis perpendicular to its length at the end. $I = \frac{1}{12}ML^2 + \frac{1}{4}MR^2 + M(L/2)^2 = \frac{1}{12}M(4R)^2 + \frac{1}{4}MR^2 + M(2R)^2 = \frac{67}{12}MR^2$.Thus,$k_2^2 = \frac{67}{12} R^2$.The ratio $\frac{k_1}{k_2} = \sqrt{\frac{2/3}{67/12}} = \sqrt{\frac{2}{3} \cdot \frac{12}{67}} = \sqrt{\frac{8}{67}}$.Therefore,$x = 67$.

List-$I$	List-$II$
$(a)$ $MI$ of the rod (length $L$,mass $M$,about an axis $\perp$ to the rod passing through the midpoint)	$(i) \frac{8ML^2}{3}$
$(b)$ $MI$ of the rod (length $L$,mass $2M$,about an axis $\perp$ to the rod passing through one of its ends)	$(ii) \frac{ML^2}{3}$
$(c)$ $MI$ of the rod (length $2L$,mass $M$,about an axis $\perp$ to the rod passing through its midpoint)	$(iii) \frac{ML^2}{12}$
$(d)$ $MI$ of the rod (length $2L$,mass $2M$,about an axis $\perp$ to the rod passing through one of its ends)	$(iv) \frac{2ML^2}{3}$

The ratio of the radius of gyration of a hollow sphere to that of a solid cylinder of equal mass,for the moment of inertia about their diameter axis $AB$ as shown in the figure,is $\sqrt{\frac{8}{x}}$. The value of $x$ is:

Similar Questions

Two rings of same mass $m$ and radius $R$ are placed with their planes perpendicular to each other and centres at a common point. The radius of gyration of the system about an axis passing through the centre and perpendicular to the plane of one ring is ............

Can the moment of inertia of the same body be different?

$ABC$ is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. $I_{AB}, I_{BC}, I_{CA}$ are the moments of inertia of the plate about axes $AB, BC,$ and $CA$ respectively. Which one of the following relations is correct?

$A$ disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. The disc is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module