A rigid body can be hinged about any point on | vedclass

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

Question

(D) According to the parallel axis theorem,the moment of inertia $I$ about an axis at position $x$ is given by $I = I_{CM} + M(x - x_{CM})^2$,where $x

Vedclass · Accepted Answer

$x = 3$

Vedclass · Answer

(D) According to the parallel axis theorem,the moment of inertia $I$ about an axis at position $x$ is given by $I = I_{CM} + M(x - x_{CM})^2$,where $x_{CM}$ is the $x$-coordinate of the centre of mass.Expanding this,we get $I = M x^2 - 2 M x_{CM} x + (I_{CM} + M x_{CM}^2)$.Comparing this with the given equation $I = 2x^2 - 12x + 27$,we equate the coefficients of $x^2$ and $x$:Coefficient of $x^2$: $M = 2$.Coefficient of $x$: $-2 M x_{CM} = -12$.Substituting $M = 2$ into the second equation: $-2(2)x_{CM} = -12 \implies -4 x_{CM} = -12 \implies x_{CM} = 3$.Thus,the $x$-coordinate of the centre of mass is $3$.

$A$ 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:

Similar Questions

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is $I_1$. The same rod is bent into a ring and its moment of inertia about the diameter is $I_2$. Then $I_1 / I_2$ is

$A$ solid sphere and a solid cylinder have the same mass and same radius. The ratio of the moment of inertia of the solid sphere about its diameter and the moment of inertia of the solid cylinder about its axis is

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is

The moment of inertia of a body does not depend on

Two rings of the same material have radii $R$ and $nR$ respectively. The ratio of their moments of inertia about an axis passing through their centers and perpendicular to their planes is $1 : 8$. The value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module