A uniform, thin rod of mass M M pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m , m, 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 ? I? Express the length L L of the rod in terms of M , M, m , m, and I .

L = √[I/[(M/2) + m]]

Step-by-step explanation:

We are given;

Mass of rod = M

Mass of each of the bodies = m

Length of rod = L

Now, moment of inertia of rod;

I_rod = (1/2)ML²

Moment of inertia of each body;

I_body = (1/2)mL²

Thus, total moment of inertia of the 3 body system is;

I = I_rod + 2(I_body)

Thus,

I = (1/2)ML² + 2((1/2)mL²)

I = (1/2)ML² + mL²

I = L²[(M/2) + m]

L² = I/[(M/2) + m]

L = √[I/[(M/2) + m]]

throw the computer

step-by-step explanation:

answer:   b

step-by-step explanation:

x2 = 16

x can not be a negative number because 16 is positive

the only answer would be positive 8