Three point masses,each of mass m,are placed | vedclass

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

Question

(C) Let the vertices of the equilateral triangle be $A$,$B$,and $C$,each with mass $m$. The axis passes through vertex $A$ and is parallel to the side

Vedclass · Accepted Answer

$\frac{3mL^2}{2}$

Vedclass · Answer

(C) Let the vertices of the equilateral triangle be $A$,$B$,and $C$,each with mass $m$. The axis passes through vertex $A$ and is parallel to the side $BC$.The moment of inertia $I$ of a system of particles is given by $I = \sum mr^2$,where $r$ is the perpendicular distance of each mass from the axis.$1$. The mass at vertex $A$ lies on the axis,so its perpendicular distance $r_A = 0$. Thus,its contribution to the moment of inertia is $m(0)^2 = 0$.$2$. The masses at vertices $B$ and $C$ are at a perpendicular distance $h$ from the axis passing through $A$. In an equilateral triangle of side $L$,the height $h$ is given by $h = L \sin 60^{\circ} = L 	imes \frac{\sqrt{3}}{2}$.$3$. The total moment of inertia $I$ is the sum of the moments of inertia of the three masses:$I = m(r_A)^2 + m(r_B)^2 + m(r_C)^2$$I = 0 + m(h)^2 + m(h)^2 = 2mh^2$Substituting the value of $h$:$I = 2m \left( L \frac{\sqrt{3}}{2} ight)^2$$I = 2m \left( \frac{3L^2}{4} ight)$$I = \frac{3mL^2}{2}$

Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $L$. The moment of inertia of the system about an axis passing through one of the vertices and parallel to the side joining the other two vertices will be

Similar Questions

The moment of inertia of a semicircular ring of mass $M$ and radius $R$ about its centre is:

We have two spheres,one of which is a hollow shell and the other solid. They have identical masses and moments of inertia about their respective diameters. The ratio of their radii is given by

The moment of inertia of a sphere of mass $M$ and radius $R$ is $I.$ If $M$ is kept constant and a graph is plotted between $I$ and $R,$ then its form would be:

The moment of inertia of a thin rod about an axis passing through its midpoint and perpendicular to the rod is $2400 \,g \,cm^2$. The length of the $400 \,g$ rod is nearly: (in $\,cm$)

Four particles each of mass $M$ are placed at the corners of a square of side $L$. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

Vedclass Test Series

Exam Paper Generator

Online Exam Module