Question

AB and CD are two identical rods each of length l and mass m joined to form a cross is fixed inside a ring of mass m and radius r l bi 2 moment of inertia of the system about a bisector of the angle between the road xy is​

Answer 1

Explanation:

2x1/12 ml2 sin245 = ml2/12 So, moment of inertia of the cross about bisector is I = ml2/12

Answer 2

Thus the moment of inertia of the cross about the bisector is I = Ml^2  / 12.

Explanation:

• Consider the line through the centre of the cross is perpendicular.
• The moment of inertia of each rod about this line is I′=Ml  2/12.
• Hence the moments of inertia of the cross I z  = 2I z ′  = Ml^ 2 /6.
• Theses moments of inertia are also equal by symmetry IX  = I Y
• According to the theorem of perpendicular axis, IZ  = I X  + I Y
• Thus the moment of inertia of the cross about the bisector is Ml^2  /12.