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 semicircular disc of mass $M$ and radius $R$ about the shown axis is:

The moment of inertia of a solid sphere,about an axis parallel to its diameter and at a distance of $x$ from it,is $I(x)$. Which one of the graphs represents the variation of $I(x)$ with $x$ correctly?

State the theorem of perpendicular axes and the theorem of parallel axes.

Find the moment of inertia of the ring shown in the figure about the axis $AB$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module