This **article aims to find the moment of inertia in the given system**. The article uses the concept of **Newton’s second law for rotational motion**.

**-Newton’s second law for rotation**, $ \sum _ { i } \tau _ { i }= I \alpha $, says that the sum of t**orques on a rotating system**Â about a fixed axis is equal to product of the moment of inertia and the**Â angular acceleration.**Â This is a**Â rotational analogy to Newton’s second law of linear motion.**

-In the vector form of **Newton’s second law for rotation**, the torque vector $ \tau $ is in the same direction as the **angular acceleration**Â $ a $. If the angular acceleration of a **rotating system is positive**, torque on the system is also **positive**, and if **angular acceleration is negative**, the torque is **negative**.

**Expert Answer**

The equivalent of **Newton’s second law for rotational motions**Â is:

\[ \tau Â = Â I Â \alpha \]

Where:

$ \tau $ is **net torque acting on the object.**

$ I $ is its **moment of inertia.**

$ \alpha $ is the**Â angular acceleration of the object**.

**Rearranging the equation**

\[ Â I Â = Â \dfrac { \tau } { \alpha } \]

And since we know the **net torque acting on the system**Â (vase+potter’s wheel), $ \tau Â = 16.0 \: Nm $, and its**Â angular acceleration**, $ \alpha Â = Â 5.69 \dfrac { rad } { s ^ { 2 } } Â $ , we can calculate the **moment of inertia of the system**:

\[ I = \dfrac { \tau } { \alpha } = \dfrac { 16.0 \: Nm } { 5.69 \: \dfrac { rad } { s ^ { 2 } } } Â = 2.81 \: kgm ^ { 2 } \]

The **moment of inertia**Â is $ 2.81 \: kgm ^ { 2 } $.

**Numerical Result**

The **moment of inertia**Â is $ 2.81 \: kgm ^ { 2 } $.

**Example**

**A clay vase on a potter’s wheel experiences an angular acceleration of $ 4 \dfrac { Â rad } { s ^ { 2 } } $ due to the application of torque of $ 10.0 \: Nm $ net. find the total moment of inertia of the vase and the potter’s wheel.**

**Solution**

The equivalent of **Newton’s second law for rotational motions**Â is:

\[ \tau Â = Â I Â \alpha \]

Where:

$ \tau $ is **net torque acting on the object**

$ I $ is its **moment of inertia**

$ \alpha $ is the**Â angular acceleration of the object**.

**Rearranging the equation:**

\[ Â I Â = Â \dfrac { \tau } { \alpha } \]

and since we know the **net torque acting on the system**Â (vase+potter’s wheel), $ \tau = 10.0 \: Nm $, and its**Â angular acceleration**, $\alpha = 4 \dfrac{ rad } { s ^ { 2 } } Â $ , we can calculate the **moment of inertia of the system**:

\[ I = Â \dfrac { \tau } { \alpha } Â = Â \dfrac { 10.0 \: Nm } { 4 \: \dfrac { rad } { s ^ { 2 } } } Â = Â 2.5 \: kgm ^ { 2 } \]

The **moment of inertia**Â is $ 2.5 \: kgm ^ { 2 } $.