Moment of inertia is a scalar. Therefore, if several objects are connected toget

Question

Moment of inertia is a scalar. Therefore, if several objects are connected together, the moment of inertia of this compound object is simply the scalar (algebraic) sum of the moments of inertia of each of the component objects. Use this principle to answer each of the following questions about the moment of inertia of compound objects:

1.A thin uniform 2.15 bar 1.20 long has a small 1.10 mass glued to each end. What is the moment of inertia of this object about an axis perpendicular to the bar through its center?

2.What is the moment of inertia of the object in part A about an axis perpendicular to the bar at one end?

3.A 700 metal wire is bent into the shape of a hoop 54.0 in diameter. Six wire spokes, each of mass 117 , are added from the center of the hoop to the rim. What is the moment of inertia of this object about an axis perpendicular to it through its center?

Explanation / Answer

first case : moment of inertia of bar is Ibar = (1/12)ML2 = (1/12)(2.15)(1.2 m)2 = 0.258 kg-m2 moment of inertia of the mass is Imass = mr2 = m(L/2)2 = (1.10 kg)(0.60 m)2 = 0.396 kg-m2 hence ,total moment of inertia is I = Ibar + 2Imass = 0.2581 kg-m2 + 2(0.396 kg-m2) = 1.05351 kg-m2 ˜ 1.053 kg-m2 b) second case : moment of inertia of bar is Ibar = (1/3)ML2 = (1/3)(2.15 kg)(1.2 m)2 = 1.032 kg-m2 moment of inertia of the mass is Imass = (1/2)mL2 = (1/2)(1.10 kg)(1.2 m)2 = 0.792 kg-m2 hence ,total moment of inertia is I = Ibar + 2Imass = 1.032 kg-m2 + 2(0.792 kg-m2) = 2.616 kg-m2 third case : mass of the metal wire M = 700 g = 0.700 kg diameter of the hoop d = 54 cm radius of the hoop r = d/2 = 27 cm = 0.27 m mass of each spoke m = 117 g =0.11 kg moment of inertia of the rim Irim = Mr2 = (0.700 kg)(0.27 m)2 = 0.0519 kg-m2 moment of inertia of the spokes Ispokes = 2mr2 = (2)(0.11 kg)(0.27 m)2 = 0.016 kg-m2 total moment of inertia of the system is I = Irim + 6Ispokes = (0.0519 kg-m2) + 6(0.016 kg-m2) = 0.148 kg-m2

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