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Formulas/physics/Rotational Motion/Rotational Kinetic Energy

Rotational Kinetic Energy

Kinetic energy of a rotating body. Rotational analogue of ½mv².
Class 11Class JEE
Derivation

The formula

KErot=12Iω2KE_{rot} = \frac{1}{2}I\omega^2

This is the kinetic energy of a body rotating about a fixed axis — the rotational analogue of 12mv2\frac{1}{2}mv^2.

Derivation

Sum the kinetic energies of all mass elements:

KErot=∑12mivi2=∑12mi(riω)2=12ω2∑miri2=12Iω2KE_{rot} = \sum \frac{1}{2}m_i v_i^2 = \sum \frac{1}{2}m_i(r_i\omega)^2 = \frac{1}{2}\omega^2 \sum m_i r_i^2 = \frac{1}{2}I\omega^2

The analogy

LinearRotational
KE=12mv2KE = \frac{1}{2}mv^2KE=12Iω2KE = \frac{1}{2}I\omega^2
KE=p22mKE = \frac{p^2}{2m}KE=L22IKE = \frac{L^2}{2I}

Both have the same structure. All linear KE results have rotational counterparts.

Work-energy theorem for rotation

Work done by a torque = change in rotational KE:

W=ΔKErot=12Iωf2−12Iωi2W = \Delta KE_{rot} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2

Example

A flywheel of I=2I = 2 kg·m² spins at ω=100\omega = 100 rad/s:

KE=12(2)(100)2=10000 J=10 kJKE = \frac{1}{2}(2)(100)^2 = 10000 \text{ J} = 10 \text{ kJ}

Flywheels store energy in this rotational form — used in hybrid vehicles and as energy buffers in power grids.

Remember
To find the torque needed to spin up a flywheel from rest to $\omega$ in time $t$: use $\alpha = \omega/t$, then $\tau = I\alpha$. Or use the work-energy approach: $W = \frac{1}{2}I\omega^2$, and $W = \tau\theta$, where $\theta = \frac{1}{2}\omega t$ for uniform angular acceleration from rest.