Four rods are arranged in the form of a squar | vedclass

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

Question

(C) The moment of inertia of a single rod $AB$ about an axis passing through its center $P$ and perpendicular to its length is $I_C = \frac{1}{12}ML^2

Vedclass · Accepted Answer

$\frac{4}{3}M{L^2}$

Vedclass · Answer

(C) The moment of inertia of a single rod $AB$ about an axis passing through its center $P$ and perpendicular to its length is $I_C = \frac{1}{12}ML^2$.Using the parallel axis theorem,the moment of inertia of rod $AB$ about an axis passing through the center $O$ of the square is:$I = I_C + Md^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}ight)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1+3}{12}ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2$.Since there are four such rods,the total moment of inertia $I'$ about the axis passing through the center $O$ and perpendicular to the plane is:$I' = 4 	imes I = 4 	imes \left(\frac{1}{3}ML^2ight) = \frac{4}{3}ML^2$.

Four rods are arranged in the form of a square. Calculate the moment of inertia about an axis passing through the center and perpendicular to the plane. (Mass $M$ and length $L$ for each rod)

Similar Questions

The moment of inertia of a ring about a diameter is

Two half rings are joined as shown in the figure. Each half ring has a radius $R$ and mass $M$. The moment of inertia of the system about the axis $XX'$ is

Five solid spheres,each of mass $m$ and radius $r$,are arranged as shown in the figure. The axis of rotation $A-A'$ passes through the centers of three spheres. Calculate the moment of inertia of the system about the axis of rotation $A-A'$. (in $m r^2$)

Two rings have their moments of inertia in the ratio $4 : 1$ and their diameters are in the ratio $4 : 1$. The ratio of their masses is

Vedclass Test Series

Exam Paper Generator

Online Exam Module