Respuesta :
Answer:
a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²
b) I = 50.39 kg m²
c) I = 16.8 kg m²
Explanation:
a) Given data:
m = 0.98 kg
a = 4.14 * 4.14
The moment of inertia is:
[tex]I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}[/tex]
For 4 particles:
[tex]I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}[/tex]
b) Distance from top left mass = x = a/2
Distance from bottom left mass = x = a/2
Distance from top right mass = x = √5 (a/2)
The total moment of inertia is:
[tex]I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}[/tex]
c)
[tex]I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}[/tex]
