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Two uniform thin identical rods each of mass M and length L are joined together to form a cross. What will be the moment of inertia of the cross about an axis passing through the point at which the two rods are joined and perpendicular to the plane of the cross?
Comments (17)
using perpendciular axis theroem.
there are two rods and MI of each is ml square by 12
since the axis pass thru the centre,moment of intertia of each rod is Ml2/12+M(l/2)2 == Ml2/3.This is for one case.Fot the other also same only
there fore total is 2*Ml2/3
using perpendciular axis theroem.
there are two rods and MI of each is ml square by 12
since the axis pass thru the centre,moment of intertia of each rod is Ml2/12+M(l/2)2 == Ml2/3.This is for one case.Fot the other also same only
there fore total is 2*Ml2/3
Cross suppose it is like alphabet x.Noe=w this axis is passing through th e point of intersection and perpendicular to the plane.Now that cross is made up of 2 rods.Take the case of one rod.Its moment of inertia is ml2/12.Now the moment of inertia for this single rod abt the given axis is found by parallel axis theorem(sorry i made a mistake earlier) axis theroem. ie ml2/12 +m(l/2)^2==ml2/3
For the other rod also MI is calculated by same way.and the total is 2*ml^2/3
Moment of Inertia: Rod
Calculating the moment of inertia of a rod about its center of mass is a good example of the need for calculus to deal with the properties of continuous mass distributions. The moment of inertia of a point mass is given by I = mr2, but the rod would have to be considered to be an infinite number of point masses, and each must be multiplied by the square of its distance from the axis. The resulting infinite sum is called an integral. The general form for the moment of inertia is:
When the mass element dm is expressed in terms of a length element dr along the rod and the sum taken over the entire length, the integral takes the form:
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Now here we assume that the two rods are forming cross such that the center of one coincides with the center of another.
Thus for the cross combination of two rods the MI will be sum of MI due to individual rods about the axis of rotation perpendicular to the rod
Thus toal Moment of inertia of the cross combiation of two rods = (1/6)ML2
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if you join n such rods with their centres coinciding and such that all lie in the same plane, irrespective of the angle between any two rods , the MI about an axis through the intersection point and perpendicular to the plane formed by the rods , each of mass m and length l, is nmL^2/12.............. please rate if you find this useful........
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