From a circular disc of radius R and mass 9M, | vedclass

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

Question

(A) The mass per unit area of the disc is $\sigma = \frac{9M}{\pi R^2}$.The mass of the removed small disc of radius $r = \frac{R}{3}$ is $m = \sigma

Vedclass · Accepted Answer

$4 MR^2$

Vedclass · Answer

(A) The mass per unit area of the disc is $\sigma = \frac{9M}{\pi R^2}$.The mass of the removed small disc of radius $r = \frac{R}{3}$ is $m = \sigma \cdot \pi r^2 = \frac{9M}{\pi R^2} \cdot \pi \left(\frac{R}{3}\right)^2 = M$.The moment of inertia $(I)$ of the complete disc about the axis passing through $O$ is $I_{total} = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$.The moment of inertia of the removed disc about its own center is $I_{cm} = \frac{1}{2}m r^2 = \frac{1}{2}M(\frac{R}{3})^2 = \frac{1}{18}MR^2$.Using the parallel axis theorem,the moment of inertia of the removed disc about the axis passing through $O$ is $I_{removed} = I_{cm} + md^2$,where $d$ is the distance between the centers,$d = R - \frac{R}{3} = \frac{2R}{3}$.$I_{removed} = \frac{1}{18}MR^2 + M(\frac{2R}{3})^2 = \frac{1}{18}MR^2 + \frac{4}{9}MR^2 = \frac{1+8}{18}MR^2 = \frac{9}{18}MR^2 = \frac{1}{2}MR^2$.The moment of inertia of the remaining disc is $I_{remaining} = I_{total} - I_{removed} = \frac{9}{2}MR^2 - \frac{1}{2}MR^2 = 4MR^2$.

From a circular disc of radius $R$ and mass $9M$,a small disc of radius $\frac{R}{3}$ is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is:

Similar Questions

From a disc of radius $R$,a concentric circular portion of radius $r$ is cut out so as to leave an annular disc of mass $M$. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its centre of gravity is

What is the moment of inertia of a ring of mass $M$ and radius $R$ about the axis $PQ$ as shown in the figure?

Three solid spheres each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about $YY'$ will be

$A$ thin wire of length $l$ and uniform linear mass density $\rho$ is bent into a circular loop with centre $O$ and radius $r$ as shown in the figure. The moment of inertia of the loop about the axis $XX'$ is

Moment of inertia of a disc of mass $M$ and radius $R$ about any of its diameter is $MR^2/4$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be $(x/2)MR^2$. The value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module