When we see a giant wheel we would be excited to see it rotating. We even enjoy sitting in it. So, if we start thinking about the physics behind giant wheel we will know that it acquires both kinetic and potential energy. Since it is moving, it possesses Kinetic energy and if it attains some height it possesses potential energy. Hence we can say that rotational energy possesses both kinetic and potential energy. We will discuss about rotational energy in this section and also about how much kinetic and potential energy it possesses.

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- Rotational Kinetic Energy
- Translational Energy

It is given by,

v = the velocity of the object or body

Similarly, rotational kinetic energy is the energy which is gained by an object or body on virtue of its rotation.

It is given by,

where,

I = moment of Inertia

$\omega$ = angular Velocity of the rotating body

The units of rotational kinetic energy = **Joules** for S.I. System

Let us derive the units by using the formula:

We know that,

K.E (rotational) = $\frac{1}{2}$ I $\omega^{2}$

Unit of Inertia: I = m r^{2}

Hence the unit for inertia is I = Kg m^{2}

Also for angular momentum $\omega$ has unit as rad/s

So, K.E (rotational) = Kg m^{2} rad^{2}/s^{2}

= watt sec

= Joules

The Mathematical formula for rotational K.E is:

As we know for translation K.E,

**K.E = $\frac{1}{2}$ mv**^{2}** **

Similarly, if we consider the rotational component then Rotational Kinetic energy is:

**K.E = $\frac{1}{2}$****I $\omega^{2}$****Rotational Kinetic Energy**.

To find the**Total Rotational Kinetic Energy**, we know that change in kinetic energy in translational motion is given by

**$\Delta$ K.E = $\frac{1}{2}$ $\Delta$ m v **^{2}

$\Delta$ K.E = $\frac{1}{2}$ $\Delta$ m r^{2} $\omega^{2}$ ..............(a)

**E**_{Total} = E _{Translational} + E _{Rotational}

**E**_{Total }**= $\frac{1}{2}$ $\Delta$ m v**^{2} + $\frac{1}{2}$ I $\omega^{2}$ ......(c)

which gives the formula for **Total Rotational Kinetic Energy**.

As we know for translation K.E,

where,

m = mass

V = velocity of the body

Similarly, if we consider the rotational component then Rotational Kinetic energy is:

where,

I = moment of inertia

$\omega$ = Angular Velocity

This is the formula for To find the

$\Delta$ K.E = $\frac{1}{2}$ $\Delta$ m r

Also

**K.E **_{total} = $\sum$ $\Delta$ K.E

= $\sum$ $\frac{1}{2}$ $\Delta$ m r^{2} $\omega^{2}$

K.E_{total}= $\omega^{2}$ $\sum$ $\frac{1}{2}$ $\Delta$ m r^{2}

= $\frac{1}{2}$ I $\omega^{2}$..............(b)

Here, I is the moment of inertia about the given axis of rotation

= $\sum$ $\frac{1}{2}$ $\Delta$ m r

K.E

= $\frac{1}{2}$ I $\omega^{2}$..............(b)

Here, I is the moment of inertia about the given axis of rotation

Potential energy is given by

where,

F = Force and

s = displacement

where,

dU = Change in potential energy

dx = Small area

Similarly, for rotational motion, we have Potential energy as,and

The law of Conservation of rotational energy states that the energy of a system remains constant.

or

Energy can neither be destroyed nor be created. It changes from one state to another. Hence for Rotational energy,

**E = K.E + P.E**Or

**m g h = $\frac{1}{2}$ m v**^{2} + $\frac{1}{2}$ m $\omega^{2}$.

where, **m**= mass of the body,

**g** = gravity,

**h** = height,

**V** = velocity of the body,

**$\omega$** = Angular Velocity.

^{}

or

Energy can neither be destroyed nor be created. It changes from one state to another. Hence for Rotational energy,

The rotational energy of the earth is an evergreen source of energy. It is an

inexhaustible resource. It will not cause any pollution, warming etc. The energy provided is 60,000,000,000 times the total electric usage of all Americans for one year.

Earth has inertia of rotation = 8.070 $\times$ 10

I = $\frac{2}{5}$ m r

m = 6 $\times$ 10

r = 6.4 $\times$ 10

It has an angular velocity of $\frac{6.2832}{86164.09}$ radians/sec approximately. It rotates once in a day.

Hence,

For a Rotating body we will use the formula:

**K.E **_{rotational} = $\frac{1}{2}$ I $\omega^{2}$.

Now, for a body the moment of inertia is given by

**I = $\frac{1}{2}$ m r**^{2}

Where, m = mass of the body

r = radius of the body

Hence,

**K.E = $\frac{1}{2}$ I $\omega^{2}$**

**K.E = $\frac{1}{2}$ ($\frac{1}{2}$ m r**^{2}) $\omega^{2}$ ........... (a)

Also 1 revolution = 2 $\pi$ rad

Hence,**$\omega$ = 2 $\pi$ $\frac{n}{t}$ ...............(b)**

where, n= no of rotations,

**K.Erotational = $\frac{1}{2}$ ($\frac{1}{2}$ mr2) $2 \pi$ $(\frac{n}{t})^{2}$ **

Now, for a body the moment of inertia is given by

Where, m = mass of the body

r = radius of the body

Hence,

Also 1 revolution = 2 $\pi$ rad

Hence,

where, n= no of rotations,

t = time taken.

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