A circular hole of radius frac{R}{4} is made | vedclass

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

Question

(B) The moment of inertia of the complete disc about point $O$ is $I_{total} = \frac{M R^2}{2}$.The radius of the removed disc is $r = \frac{R}{4}$.Si

Vedclass · Accepted Answer

$\frac{237 M R^2}{512}$

Vedclass · Answer

(B) The moment of inertia of the complete disc about point $O$ is $I_{total} = \frac{M R^2}{2}$.The radius of the removed disc is $r = \frac{R}{4}$.Since the disc is uniform,the mass is proportional to the area $(M \propto R^2)$. Therefore,the mass of the removed disc is $m = M \left( \frac{r}{R} \right)^2 = M \left( \frac{R/4}{R} \right)^2 = \frac{M}{16}$.The moment of inertia of the removed disc about its own central axis passing through $O'$ is $I_{cm} = \frac{1}{2} m r^2 = \frac{1}{2} \left( \frac{M}{16} \right) \left( \frac{R}{4} \right)^2 = \frac{M R^2}{512}$.Using the parallel axis theorem,the moment of inertia of the removed disc about point $O$ is $I_{removed} = I_{cm} + m d^2$,where $d = \frac{3R}{4}$ is the distance between $O$ and $O'$.$I_{removed} = \frac{M R^2}{512} + \left( \frac{M}{16} \right) \left( \frac{3R}{4} \right)^2 = \frac{M R^2}{512} + \frac{9 M R^2}{256} = \frac{M R^2 + 18 M R^2}{512} = \frac{19 M R^2}{512}$.The moment of inertia of the remaining portion is $I_{remaining} = I_{total} - I_{removed} = \frac{M R^2}{2} - \frac{19 M R^2}{512} = \frac{256 M R^2 - 19 M R^2}{512} = \frac{237 M R^2}{512}$.

$A$ circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass $M$ and radius $R$,as shown in the figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point $O$ and perpendicular to the plane of the disc is:

Similar Questions

$A$ sphere of diameter $r$ is cut from a sphere of radius $r$ such that the centre of mass of the remaining mass is at the maximum distance from the original centre. What is this distance?

The sector of a circular plate shown in the figure has its center of mass at $y_{CM} =$

$A$ circular plate $A$ of radius $1.5 r$ is removed from one edge of a uniform circular plate $B$ of radius $2 r$. The distance of the centre of mass of the remaining portion from the centre of the plate $B$ is

From a homogeneous square plate,a triangle is cut out as shown in the figure. The side of the square is $a$,and the apex of the triangle is at the center of the square. The distance from the center of the square to the center of mass of the remainder of the plate is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module