A uniform thin rod of mass M = 3.11 kg M=3.11 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m = 0.213 kg m=0.213 kg , are attached to the ends of the rod. What must the length L L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is I = 0.831 kg ⋅ m 2 I=0.831 kg·m2 ?

L=1.49 m

Explanation:

We are given that

We have to find the length L of the rod

Moment of inertia of the system

Hence, the length of rod=L=1.49 m

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