Two rings made of the same material and havin | vedclass

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

Question

(B) The moment of inertia $I$ of a ring about its central axis is given by $I = MR^2$.Since the rings are made of the same material with the same thic

Vedclass · Accepted Answer

$1:27$

Vedclass · Answer

(B) The moment of inertia $I$ of a ring about its central axis is given by $I = MR^2$.Since the rings are made of the same material with the same thickness $t$,the mass $M$ can be expressed as $M = 	ext{Volume} 	imes 	ext{Density} = (2\pi R \cdot A) \cdot ho$,where $A$ is the cross-sectional area and $ho$ is the density. Given the thickness is the same,$M \propto R$.Thus,$I = MR^2 \propto (R)R^2 = R^3$ is incorrect for a ring; for a ring,$M = 	ext{Area} 	imes 	ext{Circumference} 	imes ho = (t^2) 	imes (2\pi R) 	imes ho$. So $M \propto R$.Therefore,$I = MR^2 \propto R \cdot R^2 = R^3$.Wait,if the thickness $t$ is the cross-sectional radius,$M = \pi t^2 (2\pi R) ho \propto R$.Then $\frac{I_1}{I_2} = \frac{R_1^3}{R_2^3} = (\frac{0.2}{0.6})^3 = (\frac{1}{3})^3 = \frac{1}{27}$.However,if the rings are thin hoops where $M$ is proportional to the circumference $(M \propto R)$,then $I = MR^2 \propto R^3$. If the question implies the rings have the same cross-sectional area,$M \propto R$,then $I \propto R^3$. If the question implies the rings have the same mass,$I \propto R^2$. Given the standard interpretation of such problems where $M \propto R$,the ratio is $1:27$.

Two rings made of the same material and having the same thickness have radii of $0.2 \ m$ and $0.6 \ m$. The ratio of their moments of inertia is:

Similar Questions

From a circular disc of radius $R$ and mass $9M$,a small disc of mass $M$ and radius $R/3$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

Two solid spheres of iron have radii in the ratio $1 : 2$. Their moments of inertia will be in the ratio

If a solid sphere and a solid cylinder of the same radius and density rotate about their own axes,which one will have a greater moment of inertia? (Assume $L = R$)

Four equal masses,$m$ each,are placed at the corners of a square of side length $l$ as shown in the figure. The moment of inertia of the system about an axis passing through $A$ and parallel to $DB$ would be:

$A$ thin disc of mass $M$ and radius $R$ has mass per unit area $\sigma (r) = kr^2$,where $r$ is the distance from its centre. Its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module