Auniform thin rod of mass m = 4.17 kg pivots about an axis through its center and perpendicular to its length. two small bodies, each of mass m = 0.519 kg, are attached to the ends of the rod.

what must the length l of the rod be so that the moment of inertia of the three-body system with respect to the described axis is i = 0.913 kgm2?

The length of the rod must be 1.226 meter

Explanation:

Step 1: Data given

mass of rod = 4.17 kg

mass of the bodies = each 0.519 kg

length of the rod = L

I = 0.913 kg*m²

Step 2: Calculate moment of inertia of the rod

Consider L = length of the rod.

⇒ Moment of inertia of the rod = (1/12)*4.17*L²

Step 3: Moment of inertia of the 2 bodies

Moment of inertia of each of the two bodies = 0.519*(L/2)²

Step 4: Total moment of inertia

Total moment of inertia of the three-body system =

(1/12)*4.17*L² + 2*0.519*(L/2)² = 0.913 kg*m²

0.3475* L² +0.2595 *L²= 0.913

0.607 L² = 0.913

L² = 1.504

L = 1.226 m

The length of the rod must be 1.226 meter

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

spare some change and give it to the poor

explanation:

plz : ((