The moment of inertia of a uniform rod of len | vedclass

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

Question

(A) Let the rod lie along the $x$-axis with its centre at the origin. The mass per unit length of the rod is $\lambda = \frac{m}{2l}$.Consider a small

Vedclass · Accepted Answer

$\frac{m l^2}{3} \sin^2 \alpha$

Vedclass · Answer

(A) Let the rod lie along the $x$-axis with its centre at the origin. The mass per unit length of the rod is $\lambda = \frac{m}{2l}$.Consider a small element of length $dx$ at a distance $x$ from the centre. The mass of this element is $dm = \lambda dx = \frac{m}{2l} dx$.The perpendicular distance of this element from the axis $xx'$ is $r = x \sin \alpha$.The moment of inertia of this element about the axis $xx'$ is $dI = (dm) r^2 = \left(\frac{m}{2l} dx\right) (x \sin \alpha)^2$.Integrating this from $x = -l$ to $x = +l$:$I = \int_{-l}^{l} \frac{m}{2l} x^2 \sin^2 \alpha dx = \frac{m \sin^2 \alpha}{2l} \int_{-l}^{l} x^2 dx$.$I = \frac{m \sin^2 \alpha}{2l} \left[ \frac{x^3}{3} \right]_{-l}^{l} = \frac{m \sin^2 \alpha}{2l} \left( \frac{l^3}{3} - \frac{(-l)^3}{3} \right) = \frac{m \sin^2 \alpha}{2l} \left( \frac{2l^3}{3} \right) = \frac{1}{3} m l^2 \sin^2 \alpha$.

The moment of inertia of a uniform rod of length $2l$ and mass $m$ about an axis $xx'$ passing through its centre and inclined at an angle $\alpha$ is

Similar Questions

The radius of gyration of a disc of mass $50 \, g$ and radius $2.5 \, cm$ about an axis passing through its center of gravity and perpendicular to its plane is ....... $cm$.

$A$ ring,a solid sphere,and a disc have the same mass and radius. Which of them has the largest moment of inertia?

Of the two eggs which have identical sizes,shapes and weights,one is raw and the other is half-boiled. The ratio between the moment of inertia of the raw egg to the half-boiled egg about their central axis is

Three particles of mass $m$ each are placed at the three vertices of an equilateral triangle. The length of each side of the triangle is $a$. The moment of inertia of this system about any side of the triangle is:

$A$ rigid body can be hinged about any point on the $x$-axis. When it is hinged such that the hinge is at $x$,the moment of inertia is given by $I = 2x^2 - 12x + 27$. The $x$-coordinate of the centre of mass is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module