Three point masses,each of mass m,are kept at | vedclass

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

Question

(B) Let the masses be at vertices $A, B,$ and $C$ of an equilateral triangle with side $L$. Let $O$ be the centroid of the triangle.The distance from

Vedclass · Accepted Answer

$L^{3/2}$

Vedclass · Answer

(B) Let the masses be at vertices $A, B,$ and $C$ of an equilateral triangle with side $L$. Let $O$ be the centroid of the triangle.The distance from any vertex to the centroid $O$ is $R = \frac{L}{\sqrt{3}}$.The moment of inertia of the system about the axis passing through $O$ and perpendicular to the plane of the triangle is:$I = 3 	imes (m R^2) = 3 	imes m 	imes \left(\frac{L}{\sqrt{3}}ight)^2 = 3 	imes m 	imes \frac{L^2}{3} = m L^2$.For the system to rotate,the gravitational force between the masses provides the necessary centripetal force.The gravitational force on one mass due to the other two is $F_{net} = 2 	imes \left(\frac{G m^2}{L^2}ight) \cos 30^{\circ} = 2 	imes \frac{G m^2}{L^2} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3} G m^2}{L^2}$.This force acts as the centripetal force: $F_{net} = m \omega^2 R$.$\frac{\sqrt{3} G m^2}{L^2} = m \omega^2 \left(\frac{L}{\sqrt{3}}ight)$.$\omega^2 = \frac{\sqrt{3} G m}{L^2} 	imes \frac{\sqrt{3}}{L} = \frac{3 G m}{L^3}$.Since $T = \frac{2 \pi}{\omega}$,we have $T^2 = \frac{4 \pi^2}{\omega^2} = \frac{4 \pi^2 L^3}{3 G m}$.Thus,$T^2 \propto L^3$,which implies $T \propto L^{3/2}$.

Three point masses,each of mass $m$,are kept at the corners of an equilateral triangle of side $L$. The system rotates about the center of the triangle. The period of rotation $T$ is directly proportional to:

Similar Questions

For a satellite to orbit around the earth,which of the following must be true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module