Question

Four identical particles of mass 0.738 kg each are placed at the vertices of a 4.28 m x 4.28 m square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a)passes through the midpoints of opposite sides and lies in the plane of the square,(b)passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

Answer 1

given that :

mass of each particles, m = 0.738 kg

side of the square, a = 4.28 m

area of the square, A = a² = (4.28 m)²

A = 18.3 m²

the rotational inertia of this rigid body about an axis that ::

(a) passes through the midpoints of opposite sides and lies in the plane of the square whichis given as :

moment of inertia of a particles, I = m r²                 { eq. 1 }

where, r = a/2

I = m (a/2)²

or I = m a² / 4                              { eq. 2 }

there are Four identical particles which means that,

I = 4 (m a² / 4)

I = m a²                          { eq.3 }

inserting the values in eq.3

I = (0.738 kg) (4.28 m)²

I = 13.5 kg.m²

(b) passes through the midpoint of one of the sides & is perpendicular to the plane of square which is given as :

For Top left mass distance from x = a/2

For Bottom left mass distance from x = a/2

For Top right mass distance from x is given as

using a pythagoras theorem, [a