**If the axis is passing through mass A in the direction perpendicular to the page, calculate its moment of inertia with the proper unit and up to two significant figures.**

**If the axis is passing through masses B and C, calculate its moment of inertia with the proper unit and up to two significant figures.**

Figure 1

The aim of this question is to find the **Moment of Inertia** about the required **axes**.

The basic concept behind this article is the **Moment of Inertia** or **Rotational Inertia,** which is represented by the symbol $I$. It is defined as the characteristic of a **rotating body** due to which it **opposes** the **acceleration** in the **angular direction**. It is always represented in relation to an **axis of rotation**. The **Moment of Inertia** is represented by an **SI unit** of $kgm^2$ and expressed as follows:

\[I\ =\ m\ \times\ r^2\]

where,

$I=$ **Moment of Inertia**

$m=$ **Sum of the product of the mass**

$r=$ **Distance from the axis of the rotation**

## Expert Answer

Given that:

Mass $A=200g=m_1$

Mass $B=100g=m_2$

Mass $C=100g=m_3$

Distance between Mass $A\ and\ B\ =\ 10cm$

Distance between Mass $A\ and\ C\ =\ 10cm$

Distance between Mass $B\ and\ C\ =\ 12cm$

**Part-A**

**Axis** is passing **perpendicularly** through **Mass** $A$, hence we will calculate the **moment of inertia** of the system by considering **Mass** $B$ and **Mass** $C$ which are lying at a distance of $10cm$ from **Mass** $A$. As per the expression for **Moment of Inertia**, we will consider the **moment** created by both **Masses** $B$ and $C$ around the **axis** passing through **Mass** $A$ as follows:

\[I_A=m_2{r_2}^2+m_3{r_3}^2\]

Substituting the values:

\[I_A=[100g\times{(10cm)}^2]+[100g×(10cm)2]\]

\[I_A=10000g{\rm cm}^2+10000g{\rm cm}^2\]

\[I=20000g{\rm cm}^2\]

\[I_A=20000\ \frac{kg}{1000}\left(\frac{m}{100}\right)^2\]

\[I_A=2.0\ \times{10}^{-3}kgm^2\]

**Part-B**

The **axis of rotation** is passing through **Masses** B and C.

If we consider the placement of **masses** in the form of a **triangle**, the distance $r$ from **Mass** $A$ to the a**xis of rotation** will be the **height of the triangle**, and the **base** will be **half of the distance between Mass** $B$ and $C$.

Hence as per **Pythagoras’ Theorem**:

\[{\rm Hypotenuse}^2={\rm Base}^2+{\rm Height}^2\]

\[{10}^2=\left(\frac{12}{2}\right)^2+r^2\]

\[r=\sqrt{{10}^2-6^2}\]

\[r=\sqrt{64}\]

\[r=8cm\]

As per the expression for **Moment of Inertia**, we will consider the **moment** created by **Mass** $A$ around the **axis** passing through **Masses** $B$ and $C$ as follows:

\[I_{BC}=m_1r^2\]

\[I_{BC}=200g\ \times{(8cm)}^2\]

\[I_{BC}=200g\ \times{64cm}^2\]

\[I_{BC}=200g\ \times{64cm}^2\]

\[I_{BC}=12800\times\frac{kg}{1000}\left(\frac{m}{100}\right)^2\]

\[I_{BC}=1.28\times{10}^4\times{10}^{-3}\times{10}^{-4}\ kgm^2\]

\[I_{BC}=1.28\times{10}^{-3}\ kgm^2\]

## Numerical Result

**Part-A**. If the **axis** is passing through **Mass** $A$ in the **direction perpendicular** to the page, its **moment of inertia** is:

\[I_A=2.0\ \times{10}^{-3}kgm^2\]

**Part-B**. If the **axis** is passing through **Masses** $B$ and $C$, its **moment of inertia** is:

\[I_{BC}=1.28\times{10}^{-3}\ kgm^2\]

## Example

A Car having a **mass of **$1200kg$ is taking a turn around a roundabout having a **radius** of $12m$. Calculate the **moment of inertia** of the car around its roundabout.

Given that:

**Mass of the Car** $m=1200kg$

**The radius of the turn** $r=12m$

As per the expression for **Moment of Inertia**:

\[I\ =\ m\ \times\ r^2\]

\[I\ =\ 1200kg\ \times\ {(12m)}^2\]

\[I\ =\ 172800kgm^2\]

\[Moment\ of\ Inertia\ I\ =\ 1.728\times{10}^5\ kgm^2\]

*Image/Mathematical drawings are created in Geogebra.*