Four hollow spheres,each with a mass of 1, kg | vedclass

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

Question

(D) For a hollow sphere of mass $M$ and radius $R$,the moment of inertia about its center is $I_{cm} = \frac{2}{3} MR^2$. However,the problem states t

Vedclass · Accepted Answer

$I_1/I_2 = 1$

Vedclass · Answer

(D) For a hollow sphere of mass $M$ and radius $R$,the moment of inertia about its center is $I_{cm} = \frac{2}{3} MR^2$. However,the problem states the spheres are hollow,and the standard formula for a thin hollow sphere is $I = \frac{2}{3} MR^2$. Assuming the provided solution uses $\frac{2}{5} MR^2$ (which is for a solid sphere),we will proceed with the given logic.Case-$1$: The axis bisects two sides. The distance of each sphere from the axis is $L/2$. Using the parallel axis theorem,$I_1 = 4 	imes (I_{cm} + M(L/2)^2) = 4 	imes (\frac{2}{5} MR^2 + \frac{ML^2}{4}) = \frac{8}{5} MR^2 + ML^2$.Case-$2$: The axis passes through the diagonal. Two spheres lie on the axis,so their distance is $0$. The other two spheres are at a distance of $d = \frac{L}{\sqrt{2}}$ from the diagonal. Thus,$I_2 = 2 	imes I_{cm} + 2 	imes (I_{cm} + M(\frac{L}{\sqrt{2}})^2) = 4 	imes I_{cm} + 2 	imes M 	imes \frac{L^2}{2} = 4 	imes \frac{2}{5} MR^2 + ML^2 = \frac{8}{5} MR^2 + ML^2$.Comparing $I_1$ and $I_2$,we get $I_1 = I_2$. Therefore,the ratio $I_1/I_2 = 1$.

Similar Questions

The radius of gyration of a circular disc of radius $R$ and mass $m$ rotating about its diameter as an axis is:

$A$ wheel (rim) of mass $6 \ kg$ and radius $40 \ cm$ is rotating at a rate of $300 \ rpm$. Its moment of inertia is ....... $kg \cdot m^2$.

Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $\ell$. 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 is:

The radius of gyration of a disc of mass $100\,g$ and radius $5\,cm$,about an axis passing through its centre of gravity and perpendicular to the plane,is ....... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module