Four identical uniform solid spheres,each of | vedclass

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

Question

(B) The moment of inertia of a single solid sphere about its own central axis is $I_{cm} = \frac{2}{5} MR^2$. Using the parallel axis theorem,$I = I_{

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) The moment of inertia of a single solid sphere about its own central axis is $I_{cm} = \frac{2}{5} MR^2$. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance from the axis of rotation.For the axis passing through the center of sphere $A$:$I_A = I_A' + I_B' + I_C' + I_D'$$I_A = \frac{2}{5} MR^2 + (\frac{2}{5} MR^2 + M(2R)^2) + (\frac{2}{5} MR^2 + M(4R)^2) + (\frac{2}{5} MR^2 + M(6R)^2)$$I_A = \frac{8}{5} MR^2 + M(4R^2 + 16R^2 + 36R^2) = 1.6 MR^2 + 56 MR^2 = 57.6 MR^2$.For the axis passing through the center of sphere $B$:$I_B = I_A' + I_B' + I_C' + I_D'$$I_B = (\frac{2}{5} MR^2 + M(2R)^2) + \frac{2}{5} MR^2 + (\frac{2}{5} MR^2 + M(2R)^2) + (\frac{2}{5} MR^2 + M(4R)^2)$$I_B = \frac{8}{5} MR^2 + M(4R^2 + 4R^2 + 16R^2) = 1.6 MR^2 + 24 MR^2 = 25.6 MR^2$.The difference is $|I_A - I_B| = 57.6 MR^2 - 25.6 MR^2 = 32 MR^2$.

Similar Questions

The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $I$. What is the ratio $l/R$ such that the moment of inertia is minimum?

The moment of inertia of a uniform circular disc is maximum about an axis perpendicular to the disc and passing through which of the following points?

$A$ system of five solid spheres, each of mass $m$ and radius $r$, is rotating about an axis $AA'$. The moment of inertia of the system about the axis of rotation $AA'$ is: (in $mr^2$)

$A$ sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is

$A$ uniform disc of radius $R$ lies in the $x-y$ plane with its centre at the origin. Its moment of inertia about the $z$-axis is equal to its moment of inertia about the line $y = x + c$. The value of $c$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module