A thin wire of length l and mass m is bent in | vedclass

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

Question

(D) The length of the wire is $l = \pi r$,where $r$ is the radius of the semicircle.Therefore,$r = \frac{l}{\pi}$.Since the wire is a thin semicircle,

Vedclass · Accepted Answer

$\frac{2 m l^2}{\pi^2}$

Vedclass · Answer

(D) The length of the wire is $l = \pi r$,where $r$ is the radius of the semicircle.Therefore,$r = \frac{l}{\pi}$.Since the wire is a thin semicircle,every point on the wire is at a distance $r$ from the center of the semicircle.The moment of inertia of the entire wire about an axis passing through the center of the semicircle and perpendicular to its plane is $I_{center} = \int r^2 dm = r^2 \int dm = m r^2$.However,the question asks for the moment of inertia about an axis passing through the end of the wire and perpendicular to its plane.Since all mass elements $dm$ are at a distance $r$ from the center,but the axis is shifted to the end,we must consider the distance of each point from the end.For a semicircle,the moment of inertia about an axis passing through one end and perpendicular to the plane is $I = \int r^2 dm$. Since every point on the wire is at a distance $r$ from the center,but the axis is at the end,we use the parallel axis theorem or direct integration. For a thin wire bent into a semicircle,the moment of inertia about an axis perpendicular to the plane passing through one end is $I = 2mr^2$.Substituting $r = \frac{l}{\pi}$,we get $I = 2m(\frac{l}{\pi})^2 = \frac{2ml^2}{\pi^2}$.

$A$ thin wire of length $l$ and mass $m$ is bent in the form of a semicircle. The moment of inertia about an axis perpendicular to its plane and passing through the end of the wire is .............

Similar Questions

Two rods,each of mass $m$ and length $l$,are joined at their centers to form a cross. What is the moment of inertia of the cross about an axis passing through their common center and perpendicular to the plane of the rods?

$A$ one-quarter sector is cut from a uniform circular disc of radius $R$. This sector has mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

$A$ circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$,and another disc $Y$ of radius $4R$ is made from an iron plate of thickness $\frac{t}{4}$. The ratio of the moment of inertia $\frac{I_Y}{I_X}$ is:

Three particles of mass $m$ each are placed at the three vertices of an equilateral triangle. The length of each side of the triangle is $a$. The moment of inertia of this system about any side of the triangle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module