A circular disc of radius b has a hole of rad | vedclass

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

Question

(B) Consider a thin elemental ring of radius $r$ and thickness $dr$. The area of this ring is $dA = 2\pi r dr$. The mass of this element is $dm = \sig

Vedclass · Accepted Answer

$\sqrt{\frac{a^2 + b^2 + ab}{3}}$

Vedclass · Answer

(B) Consider a thin elemental ring of radius $r$ and thickness $dr$. The area of this ring is $dA = 2\pi r dr$. The mass of this element is $dm = \sigma(r) dA = \left(\frac{\sigma_0}{r}\right) (2\pi r dr) = 2\pi \sigma_0 dr$.The total mass $M$ of the disc is $M = \int_a^b dm = \int_a^b 2\pi \sigma_0 dr = 2\pi \sigma_0 (b - a)$.The moment of inertia $I$ about the central axis is $I = \int_a^b r^2 dm = \int_a^b r^2 (2\pi \sigma_0 dr) = 2\pi \sigma_0 \int_a^b r^2 dr = 2\pi \sigma_0 \left(\frac{b^3 - a^3}{3}\right)$.The radius of gyration $k$ is defined by $I = Mk^2$,so $k^2 = \frac{I}{M}$.$k^2 = \frac{2\pi \sigma_0 (b^3 - a^3) / 3}{2\pi \sigma_0 (b - a)} = \frac{b^3 - a^3}{3(b - a)}$.Using the identity $b^3 - a^3 = (b - a)(b^2 + a^2 + ab)$,we get $k^2 = \frac{(b - a)(b^2 + a^2 + ab)}{3(b - a)} = \frac{a^2 + b^2 + ab}{3}$.Therefore,$k = \sqrt{\frac{a^2 + b^2 + ab}{3}}$.

$A$ circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $\sigma(r) = \frac{\sigma_0}{r}$,then the radius of gyration of the disc about its axis passing through the centre is

Similar Questions

$A$ quarter of a uniform circular disc of radius $R$ is cut off. The mass of the cut-off part is $M$. It rotates about an axis passing through the center of the original disc and perpendicular to the plane of the disc. What will be its moment of inertia about the axis of rotation?

$A$ thin uniform wire of mass $m$ and linear density $\rho$ is bent in the form of a circular ring. The moment of inertia of the ring about a tangent parallel to its diameter is

The moment of inertia of a ring of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is:

Four masses are fixed on a massless rod as shown in the figure. The moment of inertia about the axis $PQ$ is about ..... $kg-m^2$.

Two rings have their masses in ratio $1 : 2$ and their diameters are in the ratio $2 : 1$. The ratio of their moments of inertia is

Vedclass Test Series

Exam Paper Generator

Online Exam Module