Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotational motion about a fixed axis. The formula is:
KE_rot = ½ I ω²,
where I is the moment of inertia and ω is the angular velocity. In comparison, translational kinetic energy is KE_trans = ½ m v². Moment of inertia (I) acts as the rotational analogue of mass (m), and angular velocity (ω) is the analogue of linear velocity (v).

When rolling down an incline, potential energy ($mgh$) converts into both translational and rotational kinetic energy. The split depends on the shape factor $eta$ (where $I = eta M R^2$). The hoop has $eta = 1.0$, partitioning $50%$ of its energy into rotation. The solid sphere has $eta = 0.4$, partitioning only $28.6%$ into rotation and leaving $71.4%$ for translation. Because more energy goes into translational motion, the solid sphere rolls down faster.

The Rotational Work-Energy Theorem states that the net work done by all external torques acting on a rotating body is equal to the change in the body's rotational kinetic energy:
W_rot = ∫ τ dθ = ΔKE_rot = ½ I ω_f² - ½ I ω_i².
Applying a net torque over an angular distance does rotational work, directly increasing the system's rotational kinetic energy.

No. Rotational kinetic energy ($K = rac{1}{2}Iomega^2$) and angular momentum ($L = Iomega$) both depend on the moment of inertia ($I$) and angular speed ($omega$). If a body has rotational kinetic energy ($K > 0$), it must have a non-zero spin speed ($omega > 0$). Therefore, its angular momentum must also be non-zero ($L = Iomega > 0$). Mathematically, they are related by: $K = rac{L^2}{2I}$.

A car engine generates torque in pulses (during cylinder combustion strokes). A heavy flywheel has a large moment of inertia ($I$) and stores a large amount of rotational kinetic energy at cruising speed. During the non-power strokes, the flywheel releases its stored rotational energy to keep the crankshaft rotating smoothly, dampening velocity fluctuations and smoothing engine output.