Question

Point masses of 1, 2, 3 and 10 kg are lying at the points (0, 0), (2m, 0) (0, 3m) and (- 2m, - 2m) respectively
in x-y plane. Find the moment of inertia of this system about y -axis. (in kg-m²).​

Answer 1

Answer:

The moment of inertia about y axis  $48\ \rm kg.m^2$

Explanation:

Given:

The magnitude of mass of the point masses are 1 kg 2 kg, 3 kg, 10 kg and they are placed at (0,0), (2 m,0), (0, 3 m), (-2,-2).

We know that the moment of inertia of the system of masses is given by

$I=m_1r_1^2+m_2r_2^2+m_3r_3^2+m_4r_4^2$

Where $m_1, m_2, m_3, m_4$ are the masses

$r_1, r_2, r_3, r_4$ are the distance from the axis

So According to the question we have

$I_y=1\times0^2+2\times2^2+3\times0^2+10\times2^2\\I_y=48\ \rm kg.m^2$

Answer 2

Answer:

using formula I = mr^2

The answer is 48kgm2