A square sheet of edge length L and uniform m | vedclass

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

Question

(B) The square sheet has side length $L$. When it is rolled into a hollow cylinder,the side length $L$ becomes the circumference of the cylinder's bas

Vedclass · Accepted Answer

$\frac{\sigma L^4}{4\pi^2}$

Vedclass · Answer

(B) The square sheet has side length $L$. When it is rolled into a hollow cylinder,the side length $L$ becomes the circumference of the cylinder's base.Therefore,$L = 2\pi R$,which gives the radius $R = \frac{L}{2\pi}$.The total mass $M$ of the sheet is given by $M = \sigma 	imes 	ext{Area} = \sigma L^2$.The moment of inertia $I$ of a hollow cylinder of mass $M$ and radius $R$ about its central axis is $I = MR^2$.Substituting the values of $M$ and $R$:$I = (\sigma L^2) 	imes \left(\frac{L}{2\pi}ight)^2$$I = \sigma L^2 	imes \frac{L^2}{4\pi^2}$$I = \frac{\sigma L^4}{4\pi^2}$.

$A$ square sheet of edge length $L$ and uniform mass per unit area $\sigma$ is used to form a hollow cylinder. The moment of inertia of this cylinder about its central axis is

Similar Questions

If $I_1, I_2$ and $I_3$ are moments of inertia of a solid sphere,a hollow sphere,and a ring of the same mass $M$ and radius $R$ about their respective geometrical axes,which of the following statements holds true?

The moment of inertia depends on:

If the radius of a solid sphere is $35\,cm,$ calculate the radius of gyration when the axis is along a tangent.

$A$ solid sphere $A$ is rotating about an axis $PQ$. If the radius of the sphere is $5 \, cm$ and the distance of its center from the axis $PQ$ is $10 \, cm$,then its radius of gyration about $PQ$ will be $\sqrt{x} \, cm$. The value of $x$ is $................$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module