Three uniform similar rods of length L and ma | vedclass

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

Question

(C) Let the three rods be $A$,$B$,and $C$. The axis passes through the top vertex perpendicular to the plane.For rod $A$ and rod $B$,the axis passes t

Vedclass · Accepted Answer

$\frac{L}{\sqrt{2}}$

Vedclass · Answer

(C) Let the three rods be $A$,$B$,and $C$. The axis passes through the top vertex perpendicular to the plane.For rod $A$ and rod $B$,the axis passes through one end,so their moment of inertia is $I_A = I_B = \frac{mL^2}{3}$.For rod $C$,the distance of its center of mass from the axis is $d = L \cos(30^\circ) = \frac{\sqrt{3}L}{2}$.Using the parallel axis theorem,the moment of inertia of rod $C$ about the axis is $I_C = I_{cm} + md^2 = \frac{mL^2}{12} + m\left(\frac{\sqrt{3}L}{2}\right)^2 = \frac{mL^2}{12} + \frac{3mL^2}{4} = \frac{mL^2 + 9mL^2}{12} = \frac{10mL^2}{12} = \frac{5mL^2}{6}$.The total moment of inertia $I_{total} = I_A + I_B + I_C = \frac{mL^2}{3} + \frac{mL^2}{3} + \frac{5mL^2}{6} = \frac{2mL^2 + 2mL^2 + 5mL^2}{6} = \frac{9mL^2}{6} = \frac{3mL^2}{2}$.The total mass of the system is $M = 3m$. The radius of gyration $K$ is defined by $I_{total} = MK^2$.$\frac{3mL^2}{2} = (3m)K^2 \implies K^2 = \frac{L^2}{2} \implies K = \frac{L}{\sqrt{2}}$.

Three uniform similar rods of length $L$ and mass $m$ are joined to form an equilateral triangle. Find the radius of gyration about an axis passing through one corner and perpendicular to the plane of the triangle.

Similar Questions

$A$ thin rod of mass $M$ and length $L$ is bent at its midpoint $A$ such that it forms an angle of $60^{\circ}$. What is the moment of inertia about an axis passing through the midpoint $A$ and perpendicular to the plane of the rod?

What is the moment of inertia of a cylinder of diameter $D$ and length $L$ about an axis passing through its center of gravity and perpendicular to its length?

$A$ flywheel is constructed such that its entire mass is concentrated at its rim,because......

$A$ thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^o$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

Moment of inertia of a thin uniform rod of length $L$ and mass $M$ rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent in the form of a ring,its moment of inertia about the diameter is $I_1$. If $I_1 = xI$,then the value of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module