Question

three particles of mass 10 g are placed at the vertices of an equilateral triangle of side 30 cm.find distance of its centre of mass from any of its vertex and moment of inertia of the system of particles about an axis passing through the centre of mass of the system and perpendicular to the plane containing them

Answer 1

Answer on Question #60516-Physics-Astronomy-Astrophysics

Three particles of mass $10\,\mathrm{g}$ are placed at the vertices of an equilateral triangle of side $30\,\mathrm{cm}$. Find distance of its center of mass from any of its vertex and moment of inertia of the system of particles about an axis passing through the center of mass of the system and perpendicular to the plane containing them

Solution

From symmetry, it comes that the center of mass located at the center of circumcircle of triangle, so the quantity we are looking for in part (i) is nothing else, but radius of the circumcircle.

There is simple formula of circumradius for equilateral triangle:

$$R = \frac{\sqrt{3}}{3} a$$

Where $R$ – circumradius, $a$ – side length.

Plug in numbers: $(30\,\mathrm{cm} = 0.3\,\mathrm{m})$

$$R = \frac{\sqrt{3}}{3} 0.3 = \frac{\sqrt{3}}{10} \approx 0.173\,\mathrm{m}$$

As far as we know location of the center of mass and axis, we can calculate moment of inertia:

$$I = \sum_{i} m_{i} r_{i}^{2}$$

Where $m_{i}$ – mass of the $i$th particle, $r_{i}$ – length of perpendicular from the $i$th particle to the axis passing through the center of mass.

In our case all masses and all distances are equal, that’s why we can make simplification:

$$I = \sum_{i} m_{i} r_{i}^{2} = m r^{2} \sum_{i} 1 = 3 m r^{2}$$

Plug in numbers: $(10\,\mathrm{g} = 0.01\,\mathrm{kg})$

$$I = 3 \cdot 0.01 \cdot \left(\frac{\sqrt{3}}{10}\right)^{2} = 9 \cdot 10^{-4}\,\mathrm{kg} \cdot \mathrm{m}^{2}.$$

https://www.AssignmentExpert.com