From a disc of radius R,a concentric circular | vedclass

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

Question

(A) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc of radius $R$ is $M_1 = \sigma \pi R^2$ and its moment of iner

Vedclass · Accepted Answer

$ \frac{1}{2}M(R^2 + r^2) $

Vedclass · Answer

(A) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc of radius $R$ is $M_1 = \sigma \pi R^2$ and its moment of inertia is $I_1 = \frac{1}{2} M_1 R^2 = \frac{1}{2} (\sigma \pi R^2) R^2 = \frac{1}{2} \sigma \pi R^4$.The mass of the removed circular portion of radius $r$ is $M_2 = \sigma \pi r^2$ and its moment of inertia is $I_2 = \frac{1}{2} M_2 r^2 = \frac{1}{2} (\sigma \pi r^2) r^2 = \frac{1}{2} \sigma \pi r^4$.The mass of the remaining annular disc is $M = M_1 - M_2 = \sigma \pi (R^2 - r^2)$,which implies $\sigma \pi = \frac{M}{R^2 - r^2}$.The moment of inertia of the annular disc is $I = I_1 - I_2 = \frac{1}{2} \sigma \pi (R^4 - r^4)$.Substituting $\sigma \pi$,we get $I = \frac{1}{2} \left( \frac{M}{R^2 - r^2} \right) (R^4 - r^4) = \frac{1}{2} \left( \frac{M}{R^2 - r^2} \right) (R^2 - r^2)(R^2 + r^2) = \frac{1}{2} M(R^2 + r^2)$.

From a disc of radius $R$,a concentric circular portion of radius $r$ is cut out so as to leave an annular disc of mass $M$. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is

Similar Questions

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is:

Which one of the following is not a biopolymer?

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides of the triangle are in

Observe the following statements regarding the purification of bauxite:
$I$. During Hall's process,silica is removed as $Si$ (vapour).
$II$. Bauxite ore contaminated with $Fe_2O_3$ is purified in Baeyer's process.
$III$. During Serpeck's process,$AlN$ is formed.
The correct answer is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module