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

Answer:

$I = 0.00801 kg.m^2$

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:

$ma = \frac{La*m}{La+Lb}=0.00802kg$

$mb = \frac{Lb*m}{La+Lb}=0.0125kg$

The total inertia is given by:

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

The inertia is therefore:

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

$I = 0.00801 kg.m^2$