Rotational Kinetic Energy Derivation

To derive the formula of rotational kinetic energy consider the kinetic energy of individual particles within a rotating object, Let us assume that:

• m₁, m₂, …, m_n are the masses of n particles in a rigid body rotating about an axis.
• r₁, r₂ ,…, r_n are their distances from the axis of rotation respectively with angular velocity ω.

We know that,

The kinetic energy of an object is given by, K.E.=1/2mv²

And the relation between linear velocity and angular velocity is, v = rω

Therefore, rotational kinetic energy becomes, K.E. = 1/2mr²ω²

The rotational kinetic energy (K.E.) of each particle can be written as:

• Rotational K.E. of the first particle, E₁ = 1/2m₁r₁²ω²

• Rotational K.E. of the second particle. E₂ = 1/2m₂r₂²ω²

• Similarly, Rotational K.E. of n^th particle will be E_n = 1/2m_nr_n²ω²

Hence the total Rotational kinetic energy of all the particles becomes:

E = E₁ + E₂ + ….. + E_n

Substitute the value of E₁ , E₂ ,…, E_n we get:

E = [1/2m₁r₁²ω² + 1/2m₂r₂²ω² + … + 1/2m_nr_n²ω²]

On simplifying it we get:

E = 1/2[m₁r₁²+m₂r₂²+…+m_nr_n²]ω²

E = 1/2 [[Tex]\sum_{i = 1}^{n}[/Tex]m_ir_i²] ω²

We know, the moment of inertia I = [Tex]\sum_{i = 1}^{n}[/Tex]m_ir_i², Hence our equation becomes:

E = 1/2 I⋅ω²

Therefore, the rotational kinetic energy of an object with moment of inertia (I) and angular velocity (ω) is given by E = 1/2 I⋅ω²

Rotational Kinetic Energy of Solid Sphere

By using the kinetic energy formula, E = 1/2 Iω² rotational kinetic energy of solid sphere can be calculated.

Solid sphere rotating about its axis has moment of inertia is, I = 2/5mR²

where m and R are mass and radius of solid sphere.

Therefore, by substituting value of I in the formula of rotational kinetic energy we get:

K = 1/2(2/5mR²)ω²

K = 1/5mR²ω²

Hence, the rotational Kinetic Energy of Solid Sphere is K = 1/5mR²ω²

Rotational Kinetic Energy of Earth

By using Rotational Kinetic Energy formula, E=1/2 Iω² we can be calculate, rotational kinetic energy of Earth.

Earth’s moment of inertia is 8 × 10³⁷ kgm²(approx.).

Its average angular velocity of Earth is 7.27×10^-5 radians per second(approx.)

Putting these value in the formula of rotational kinetic energy we get:

E= 1/2 × 8×1037 kg m² × (7.27×10^-5 rad/s)²

E= 1/2 × 8×1037 kg m² × 5.29 × 10 ⁻⁹ kg m² /s²

K = 2.12×10²⁹ Joules

Hence, Earth’s rotational kinetic energy is approximately equal to 2.12×10²⁹ Joules.

Rotational Kinetic Energy of Disc

By using the kinetic energy formula, E = 1/2 Iω² rotational kinetic energy of disc can be calculated.

Disc rotating about its axis, has moment of inertia, I = 1/2mR²

where m and R are mass and radius of disc.

Therefore, by substituting the value of I we got the expression as,

K = 1/2(1/2mR²)ω²

K = 1/4mR²ω²

Hence, the rotational Kinetic Energy of a Disc is K = 1/4mR²ω²

Rotational Kinetic Energy is described as the kinetic energy associated with the rotation of an object around an axis. It is also known as angular kinetic energy. It is dependent on the mass of an object and its angular velocity.

In this article, we will learn about rotational kinetic energy, its formula and derivation, examples of rotational kinetic energy, and the difference between rotational and translational kinetic energy.