# A clay vase on a potter’s wheel experiences an angular acceleration of 5.69 rad/s^2 due to the application of a 16.0 nm net torque. find the total moment of inertia of the vase and potter’s wheel.

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 torques on a rotating system about a fixed axis is equal to product of the moment of inertia and the angular acceleration. This is a 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 $a$. 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.

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

$\tau = I \alpha$

Where:

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

$I$ is its moment of inertia.

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

Rearranging the equation

$I = \dfrac { \tau } { \alpha }$

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

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

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

## Numerical Result

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

## Example

A clay vase on a potter’s wheel experiences an angular acceleration of $4 \dfrac { rad } { s ^ { 2 } }$ 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 is:

$\tau = I \alpha$

Where:

$\tau$ is net torque acting on the object

$I$ is its moment of inertia

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

Rearranging the equation:

$I = \dfrac { \tau } { \alpha }$

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

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

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