The moment of inertia of a solid sphere about | vedclass

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

Question

(D) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I = \frac{2}{5}MR^2$. Let the radius of the lar

Vedclass · Accepted Answer

$\frac{I}{243}$

Vedclass · Answer

(D) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I = \frac{2}{5}MR^2$. Let the radius of the large sphere be $R$ and its mass be $M$. When it is recast into $27$ small spheres of radius $r$,the volume remains constant: $\frac{4}{3}\pi R^3 = 27 	imes \frac{4}{3}\pi r^3$. This implies $R^3 = 27r^3$,so $R = 3r$ or $r = \frac{R}{3}$. The mass of each small sphere $m$ is $\frac{M}{27}$. The moment of inertia of each small sphere about its diameter is $I' = \frac{2}{5}mr^2$. Substituting $m = \frac{M}{27}$ and $r = \frac{R}{3}$: $I' = \frac{2}{5} \left( \frac{M}{27} ight) \left( \frac{R}{3} ight)^2 = \frac{2}{5} \left( \frac{M}{27} ight) \left( \frac{R^2}{9} ight) = \frac{1}{243} \left( \frac{2}{5}MR^2 ight)$. Since $I = \frac{2}{5}MR^2$,we get $I' = \frac{I}{243}$.

The moment of inertia of a solid sphere about its diameter is $I$. It is then recast into $27$ small spheres of the same diameter. The moment of inertia of each small sphere about its diameter is:

Similar Questions

$A$ disc $X$ of radius $R$ and thickness $t$ is made of an iron plate. Another disc $Y$ of radius $4R$ and thickness $t/4$ is also made of the same iron plate. What is the relationship between their moments of inertia $I_x$ and $I_y$?

What is analogous to mass in linear motion for rotational motion?

$A$ 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

Find the moment of inertia of the system of particles about the axis $OA$ passing through point $O$.

Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ and moments of inertia $I_1$ and $I_2$,respectively. For $R_2 = 2 R_1$,the ratio of $I_1$ and $I_2$ would be $1 / x$,where $x = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module