Two uniform thin identical rods AB and CD eac | vedclass

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

Question

(C) Let the intersection point of the rods be the origin $(0,0)$.For rod $AB$ lying along the $x$-axis,the moment of inertia about the $x$-axis is $I_

Vedclass · Accepted Answer

$\frac{ML^2}{12}$

Vedclass · Answer

(C) Let the intersection point of the rods be the origin $(0,0)$.For rod $AB$ lying along the $x$-axis,the moment of inertia about the $x$-axis is $I_x = 0$ and about the $y$-axis is $I_y = \frac{ML^2}{12}$.For rod $CD$ lying along the $y$-axis,the moment of inertia about the $x$-axis is $I_x = \frac{ML^2}{12}$ and about the $y$-axis is $I_y = 0$.The total moment of inertia of the cross about the $x$-axis is $I_x = 0 + \frac{ML^2}{12} = \frac{ML^2}{12}$.The total moment of inertia of the cross about the $y$-axis is $I_y = \frac{ML^2}{12} + 0 = \frac{ML^2}{12}$.The line $EF$ is a bisector in the $xy$-plane,making an angle of $45^{\circ}$ with both axes.The moment of inertia about an axis in the plane of the object is given by $I = I_x \sin^2 	heta + I_y \cos^2 	heta - 2 I_{xy} \sin 	heta \cos 	heta$.Since the rods are symmetric about the axes,the product of inertia $I_{xy} = 0$.Thus,$I_{EF} = I_x \sin^2 45^{\circ} + I_y \cos^2 45^{\circ} = \frac{ML^2}{12} \left( \frac{1}{2} ight) + \frac{ML^2}{12} \left( \frac{1}{2} ight) = \frac{ML^2}{12}$.

Two uniform thin identical rods $AB$ and $CD$ each of mass $M$ and length $L$ are joined so as to form a cross as shown. The moment of inertia of the cross about a bisector line $EF$ is (Line $EF$ is in the plane of the cross and bisects the angle between the rods).

Similar Questions

An annular ring has mass $10 \ kg$ and inner and outer radii are $5 \ m$ and $10 \ m$ respectively. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is

The analogue of mass in rotational motion is:

Radius of gyration of a thin uniform circular disc about the axis passing through its centre and perpendicular to its plane is $K_{c}$. Radius of gyration of the same disc about a diameter of the disc is $K_d$. The ratio $K_c: K_d$ is

$A$ swimmer curls their body before diving into the water from a height so that:

Three identical thin rods each of mass $m$ and length $l$ are placed along $x, y$ and $z$-axes respectively. They are placed such that one end of each rod is at the origin $O$. The moment of inertia of this system about the $z$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module