Two discs of same mass and same thickness (t) | vedclass

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

Question

(B) The moment of inertia of a disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = \frac{1

Vedclass · Accepted Answer

$d_2 : d_1$

Vedclass · Answer

(B) The moment of inertia of a disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = \frac{1}{2} M R^2$.Since the mass $M$ is the same for both discs,the ratio is $\frac{I_1}{I_2} = \frac{R_1^2}{R_2^2}$.The mass of a disc is given by $M = 	ext{Volume} 	imes 	ext{Density} = (\pi R^2 t) d$.Since $M$ and $t$ are constant,we have $R^2 \propto \frac{1}{d}$,which implies $R^2 d = 	ext{constant}$.Therefore,$R_1^2 d_1 = R_2^2 d_2$,which gives $\frac{R_1^2}{R_2^2} = \frac{d_2}{d_1}$.Substituting this into the ratio of moments of inertia,we get $\frac{I_1}{I_2} = \frac{d_2}{d_1}$.

Two discs of same mass and same thickness $(t)$ are made from two different materials of densities $d_1$ and $d_2$ respectively. The ratio of the moment of inertia $I_1$ to $I_2$ of the two discs about an axis passing through the centre and perpendicular to the plane of the disc is:

Similar Questions

The radius of gyration of a circular disc of radius $R$ and mass $m$ rotating about its diameter as an axis 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:

Two rings of radius $R$ and $nR$ having different masses and made up of the same wire have the ratio of moment of inertia about an axis passing through the centre as $1 : 8$. The value of $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module