Question

4 particles of mass $ 2kg $ , $ 3kg $ , $ 5kg $ and $ 2kg $ are situated at points A, B, C, and D respectively of a square ABCD of sides $ 1m $ respectively. Find the Moment of Inertia of the square about an axis through the center of the square and perpendicular to the plane of the square.

Answer 1

Hint To solve this question, we have to calculate the moment of inertia about the given axis due to each of the four masses and then sum all of them to get the final moment of inertia about the center.

Formula Used: The formula used to solve this question is given as,
 $\Rightarrow I = m{x^2} $
Here, $ I $ is the moment of Inertia, $ m $ is the mass of each particle, $ x $ is the perpendicular distance of the particle from the axis.

Complete step by step answer
We start by calculating the moment of inertia of each particle about the given axis.
Since, all the sides of the square are equal, so the distance of all the particles from the center shall be equal.
Let the side of the square be $ a $ .

Using Pythagoras theorem in the triangle AOB, we have
$\Rightarrow O{A^2} + O{B^2} = A{B^2} $
$\Rightarrow 2{\left( {OA} \right)^2} = {a^2} $
This gives us,
 $\Rightarrow OA = \dfrac{a}{{\sqrt 2 }} $
Hence the distance between the particle at the edge and the center of the square is $ \dfrac{a}{{\sqrt 2 }} $
Now we calculate the moment of inertia by using
$\Rightarrow I = \sum\limits_{}^{} {m{x^2}} $
$\Rightarrow I = (2){(\dfrac{a}{{\sqrt 2 }})^2} + (3){(\dfrac{a}{{\sqrt 2 }})^2} + (5){(\dfrac{a}{{\sqrt 2 }})^2} + (2){(\dfrac{a}{{\sqrt 2 }})^2} $
Since, length of the square is 1m, so we get,
 $\Rightarrow I = (2)(\dfrac{1}{2}) + (3)(\dfrac{1}{2}) + (5)(\dfrac{1}{2}) + (2)(\dfrac{1}{2}) $
 $ \Rightarrow I = 1 + \dfrac{3}{2} + \dfrac{5}{2} + 1 $
This gives us,
 $\Rightarrow I = 2 + \dfrac{{3 + 5}}{2} $
 $\Rightarrow I = 2 + 4 $
This gives,
 $\Rightarrow I = 6Kg{m^2} $
$\Rightarrow \therefore $ 6 Kgm2 is the moment of inertia of the square about an axis perpendicular to the plane of the sphere and passing through the center of the square.

Note
While solving these types of questions, always try to draw the figure corresponding to the problem. This is because in these types of questions, most of the time there is a symmetry which simplifies the calculations and reduces our time and efforts. The symmetry will be visible only when we have the corresponding figure.