Question

A thin uniform wire is bent into a rectangle. The short, vertical sides are of length a, and the long, horizontal sides are of length b.

If the total mass is 41.00 grams, a = 45.00 cm and b = 70.00 cm, what is the moment of rotational inertia about an axis through one of the vertical wires?

Answer 1

`I = 0.00801 kg.m^2`

Step-by-step explanation:

The rectangle is composed by 4 pieces:

2*ma + 2*mb = 0.041 Kg Since the wire is uniform, we can calculate each partial mass ma and mb:

`ma = (La*m)/(La+Lb)=0.00802kg`

`mb = (Lb*m)/(La+Lb)=0.0125kg`

The total inertia is given by:

`I=2*Ib+ma*Lb^2` where
`Ib=mb/3*Lb^2`

The inertia is therefore:

`I=2*mb/3*Lb^2+ma*Lb^2`

`I = 0.00801 kg.m^2`