Answer:
a) [tex]I=3.75\ kg.m^2[/tex]
b) [tex]I=11.25\ kg.m^2[/tex]
Explanation:
Given:
mass of four identical particles placed on the vertices of 2.5 m square, [tex]m=0.6\ kg[/tex]
a)
Now, when the square frame is mass-less and the axis of rotation passes through the plane of the square from the midpoints of the opposite sides then we have 4 point masses rotating about the axis.
So, rotational inertia of the system:
[tex]I=4\times [m\times \frac{2.5}{2}][/tex]
[tex]I=4\times 0.6\times (\frac{2.5}{2})^2[/tex]
[tex]I=3.75\ kg.m^2[/tex]
b)
When the axis of rotation passes through the midpoint of one of the sides of square and is perpendicular to the plane of the square we get one pair at a certain distance and the other pair at some other certain distance as shown in the schematic.
[tex]I=2\times (0.6\times 1.25^2)+2\times (0.6\times 2.795^2)[/tex]
[tex]I=11.25\ kg.m^2[/tex]
Answer
given,
mass of the four identical particle , m= 0.60 kg
dimension of the square = 2.5 m x 2.5 m
a) Rotational inertia when axis is passing through midpoints of opposite sides and lies in the plane of the square.
m₁ = m₂ = m₃ = m₄ = m
r₁ = r₂ = r₃ = r₄ = (a/2) = 2.5/2 = 1.25 m where a is the side of the square
I = m₁r₁² + m₂r₂² + m₃r₃² +m₄r₄²
I = 4 x m (a/2)²
I = 4 x 0.6 x 1.25²
I = 3.75 kg.m²
b) now,
m₁ = m₂ = m₃ = m₄ = m
I = m₁r₁² + m₂r₂² + m₃r₃² +m₄r₄²
I = [tex]0.6 \times (\dfrac{a}{2})^2 +0.6 \times (\dfrac{a}{2})^2 + 0.6 \times ((\dfrac{a}{2})^2+a^2) + 0.6 \times ((\dfrac{a}{2})^2+ a^2)[/tex]
I = [tex]0.6 \times (\dfrac{2.5}{2})^2 +0.6 \times (\dfrac{2.5}{2})^2 + 0.6 \times ((\dfrac{2.5}{2})^2+2.5^2) + 0.6 \times ((\dfrac{2.5}{2})^2+ 2.5^2)[/tex]
I = 11.25 kg.m²
