Question

A system of four particles made of 20 g at (4,0), 30 g at (0,6), 25 g at (-4,3) and 40 g at (-3,2) in the xy-plane, rotate about the z-axis through the centre of the coordinate system. The coordinates are given in units of centimetres. Calculate the moment of inertia of the system.

a. 2545gcm2
b. 1850gcm2
c. 3234gcm2
d. 2545gcm2

Answer 1

Answer on Question#41436 – Physics – Mechanics | Kinematics | Dynamics

Moment of inertia of the system: $I = \sum_{i=1}^{4} I_i$

Where $I_i$ is moment of inertia of one particle: $I_i = m_i \cdot R^2$

$m_i$ – mass of the particle; $R$ – the distance to the particle

$$I_1 = m_1 \cdot R_1^2 = 20 \cdot \left(\sqrt{4^2 + 0^2}\right)^2 = 20 \cdot 16 = 320 (g \cdot cm^2)$$ $$I_2 = m_2 \cdot R_2^2 = 30 \cdot \left(\sqrt{0^2 + 6^2}\right)^2 = 30 \cdot 36 = 1080 (g \cdot cm^2)$$ $$I_3 = m_3 \cdot R_3^2 = 25 \cdot \left(\sqrt{(-4)^2 + 3^2}\right)^2 = 25 \cdot 25 = 625 (g \cdot cm^2)$$ $$I_4 = m_4 \cdot R_4^2 = 40 \cdot \left(\sqrt{(-3)^2 + 2^2}\right)^2 = 40 \cdot 13 = 520 (g \cdot cm^2)$$

**Answer:**

Moment of inertia of the system: $I = \sum_{i=1}^{4} I_i = 320 + 1080 + 625 + 520 = 2545 (g \cdot cm^2)$

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