# Rotational Kinetic Energy Formula

The power of moving entities is referred to as kinetic energy. It might be either longitudinal or rotating. As a result, an entity in the rotation will also have kinetic energy. Rotational kinetic energy is the name given to this type of energy. Such energy is determined by its angular acceleration, which is measured in radians per second. It also relies on the pivot point of the object. Deflection is a measurement of how simple it can be to alter an object’s rotating state.

### Rotational Kinetic Energy

The energy released by an entity or object as a result of its rotation is known as rotational kinetic energy. The work-energy concept can be used to express the formulas for longitudinal and rotational inertia energy. Consider the following analogy: a constant force applied to a spinning wheel with a moment of inertia I and a constant force applied to a mass m, both starting at rest. Here, the work done would be = τ × θ, if Newton’s second law of motion were to be applied.

**Rotational Kinetic Energy Formula**

E_{k}= ½ × I × ω^{2}Where,

- I denotes the inertia moment.
- ω denotes the angular velocity.

**Derivation**

Some kind of force is required to spin items like shards. When such force acts perpendicular to the circle of a disc and remains opposite as the disc rotates, this is the most basic spinning example. So the net work done here is given as follows:

W = F × Δs

⇒ W = F × Δs × r/r

Since, r net F = net F τ and Δs/r = θ

W = τ × θ

This equation is the formula for rotational work. It’s a lot like the common standard of longitudinal work, that is product of force and distance. Angle is equivalent to distance, while torque is equivalent to force. Despite the fact that it was created for a specific case, the equation is applicable to any situation. Now, since τ = Iα, we have:

Net W = I × α × θ

⇒ ω

^{2 }= ω_{o}^{2}+ 2αθ⇒ W = ½ Iω

^{2}– ½ Iω_{o}^{2}

⇒ E_{k}= ½ × I × ω^{2}Hence proved.

### Sample Problems

**Problem 1: Find the rotational kinetic energy of an object given its angular velocity is 80π rad/sec and I = 25 kg m ^{2}.**

**Solution:**

Given: I = 25 kg m

^{2}and ω = 80π rad/secSince, E

_{k}= ½ × I × ω^{2}= ½ × 25 × 80π

= 3140 J

**Problem 2: Find the rotational kinetic energy of an object given its angular velocity is 90π rad/sec and I = 20 kg m ^{2}.**

**Solution:**

Given: I = 20 kg m

^{2}and ω = 90π rad/secSince, E

_{k}= ½ × I × ω^{2}= ½ × 20 × 90π

= 2826 J

**Problem 3: Find the angular velocity if rotational kinetic energy of an object is 1648.5 J and I = 15 kg m ^{2}.**

**Solution:**

Given: E

_{k}= 1648.5 J and I = 15 kg m^{2}.Since, E

_{k}= ½ × I × ω^{2}=> ω

^{2}= 2E_{k}/I=> ω

^{2}= 2 (1648.5)/15=> ω = 14.82 rad/sec

**Problem 4: Find the angular velocity if rotational kinetic energy of an object is 14506.8 J and I = 77 kg m ^{2}.**

**Solution:**

Given: E

_{k}= 14506.8 J and I = 77 kg m^{2}.Since, E

_{k}= ½ × I × ω^{2}=> ω

^{2}= 2E_{k}/I=> ω

^{2}= 2 (14506.8)/77=> ω = 19.41 rad/sec

**Problem 5: Find the moment of inertia if rotational kinetic energy of an object is 1884 J and its angular velocity is 40π rad/sec.**

**Solution:**

Given: E

_{k}= 1884 J and ω = 30π rad/secSince, E

_{k}= ½ × I × ω^{2}=> I = 2E

_{k}/ω^{2}= 2 (1884)/(30π)

^{2}= 35 kg m

^{2}

**Problem 6: Find the moment of inertia if rotational kinetic energy of an object is 138164 J and its angular velocity is 80π rad/sec.**

**Solution:**

Given: E

_{k}= 13816 J and ω = 80π rad/secSince, E

_{k}= ½ × I × ω^{2}=> I = 2E

_{k}/ω^{2}= 2 (13816)/(80π)

^{2}= 55 kg m

^{2}

**Problem 7: Find the rotational kinetic energy of an object given its angular velocity is 10π rad/sec and I = 6 kg m ^{2}.**

**Solution:**

Given: I = 6 kg m

^{2 }and ω = 10π rad/secSince, E

_{k}= ½ × I × ω^{2}= ½ × 6 × 10π

= 94.2 J