Two identical spherical balls of mass M and r | vedclass

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

Question

(A) The system consists of a rod of mass $M$ and length $L = 2R$,and two spheres of mass $M$ and radius $R$ attached at the ends.$1$. Moment of inerti

Vedclass · Accepted Answer

$\frac{137}{15}MR^2$

Vedclass · Answer

(A) The system consists of a rod of mass $M$ and length $L = 2R$,and two spheres of mass $M$ and radius $R$ attached at the ends.$1$. Moment of inertia of the rod about an axis passing through its center and perpendicular to its length is $I_{	ext{rod}} = \frac{ML^2}{12} = \frac{M(2R)^2}{12} = \frac{4MR^2}{12} = \frac{1}{3}MR^2$.$2$. For each sphere,the distance from the center of the rod to the center of the sphere is $d = R + \frac{L}{2} = R + R = 2R$.$3$. Using the parallel axis theorem,the moment of inertia of one sphere about the axis is $I_{	ext{sphere}} = I_{	ext{cm}} + Md^2 = \frac{2}{5}MR^2 + M(2R)^2 = \frac{2}{5}MR^2 + 4MR^2 = \frac{22}{5}MR^2$.$4$. Total moment of inertia $I = I_{	ext{rod}} + 2 	imes I_{	ext{sphere}} = \frac{1}{3}MR^2 + 2 	imes \left( \frac{22}{5}MR^2 ight) = \frac{1}{3}MR^2 + \frac{44}{5}MR^2$.$5$. Taking the common denominator $15$,$I = \left( \frac{5 + 132}{15} ight) MR^2 = \frac{137}{15}MR^2$.

Two identical spherical balls of mass $M$ and radius $R$ each are stuck on two ends of a rod of length $2R$ and mass $M$ (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

Similar Questions

Consider a thin uniform square sheet made of a rigid material. If its side is $a$,mass is $m$,and the moment of inertia about one of its diagonals is $I$,then:

The moment of inertia of a triangular plate $ABC$ of mass $M$ and side $BC = a,$ about an axis passing through $A$ and perpendicular to the plane of the plate is :-

The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through a point at a distance $L/4$ from one of its ends and perpendicular to the length of the rod is

Vedclass Test Series

Exam Paper Generator

Online Exam Module