From a uniform circular thin disc of mass 9 M | vedclass

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

Question

(A) The mass of the original disc is $M_{total} = 9M$ and its radius is $R$. The moment of inertia of the complete disc about an axis passing through

Vedclass · Accepted Answer

$4 MR^{2}$

Vedclass · Answer

(A) The mass of the original disc is $M_{total} = 9M$ and its radius is $R$. The moment of inertia of the complete disc about an axis passing through its centre and perpendicular to its plane is $I_{1} = \frac{1}{2} M_{total} R^{2} = \frac{1}{2} (9M) R^{2} = \frac{9 MR^{2}}{2}$.Since the mass is proportional to the area,the mass of the removed disc $m$ is given by $m = M_{total} 	imes \frac{\pi (R/3)^{2}}{\pi R^{2}} = 9M 	imes \frac{1}{9} = M$.The moment of inertia of the removed disc about the same axis is calculated using the parallel axis theorem: $I^{\prime} = I_{cm} + m d^{2}$,where $I_{cm} = \frac{1}{2} m (R/3)^{2}$ and $d = \frac{2R}{3}$.$I^{\prime} = \frac{1}{2} M (R/3)^{2} + M (2R/3)^{2} = \frac{MR^{2}}{18} + \frac{4MR^{2}}{9} = \frac{MR^{2} + 8MR^{2}}{18} = \frac{9MR^{2}}{18} = \frac{MR^{2}}{2}$.The moment of inertia of the remaining disc is $I_{remaining} = I_{1} - I^{\prime} = \frac{9 MR^{2}}{2} - \frac{MR^{2}}{2} = \frac{8 MR^{2}}{2} = 4 MR^{2}$.

Similar Questions

Four spheres of diameter $2a$ and mass $M$ are placed with their centres on the four corners of a square of side $b$. The moment of inertia of the system about an axis passing through one of the sides of the square is ..........

Consider a uniform square plate of side '$a$' and mass '$m$'. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

$A$ solid sphere of radius $R$ and mass $M$ is rotating about its diameter. The moment of inertia of the solid sphere rotating about an axis at a distance $R/3$ from the centre and parallel to that diameter is

The moment of inertia of a uniform hollow hemisphere about the given axes $I_A$ and $I_B$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module