Question

Four identical particles of mass m each are placed at the vertices of a square with side length a and held by four massless rods, which form the sides of the square. (Use any variable stated above.)  

a. What is the rotational inertia of this rigid body about an axis passes through the midpoint of opposite sides and lies in the plane of the square? 

b. What is the rotational inertia of this rigid body about an axis that passes through the midpoint of one of the sides and perpendicular to the plane of the square?

c. What is the rotational inertia of this rigid body about an axis that lies in the plane of the square and passes through two diagonally opposite particles? 

Answer 1

(1) The axis passes through the centre of the square and the midpoints of two opposite sides. Hence,

I = 4xm(a/2)^2 (since the distance to each mass is a/2).

implies, I = ma^2.

(2) The axis passes through the midpoint of one of the sides and is perpendicular to the plane. Hence,

I = 2xm(a/2)^2 + 2xmx5a^(2)/4 (for the two particles on the side through whose midpoint the axis passes the implies, I = 3ma^2. distance is a/2 + for the other two particles it is sqrt( (a/2)^2+a^2 ) ).

(3) The axis passes through two diagonally opposite particles. Hence the moment of inertia of the two particles about this axis will be zerosince they lie on the axis itself. The only contributing particles will be the other two. Hence,

I = 2xmxa^(2)/2 (for the other two particles the distance is a/sqrt(2) ).

implies, I = ma^2.

Answer 2

a) I = 4 * m (1/2 a)^2 = ma^2, since the distance to each mass is 1/2 a

b) I = 2 * m (1/2 a)^2 + 2 * m ( sqrt ( (a/2)^2 + a^2) ) ^2 = 1/2 ma^2 + 2m (5/4 a^2) = 1/2 ma^2 + 5/2 ma^2 = 3 ma^2

c) I = 4 * m (sqrt(2) a / 2)^2 = 2 ma^2