The quantities of angular motion (angular displacement, velocity and acceleration), moment of inertia, angular momentum and its conservation, and how a performer controls rotation in flight.

A Component 01 Section C calculation. One mark for the equation, one for the value, one for the unit.

Use angular momentum equals moment of inertia times angular velocity, $H = I \times \omega$, so $H = 12 \times 5 = 60$. The unit is kilograms metres squared radians per second (kg m squared rad/s), often written kg m squared per second in PE.

A common error is to add or subtract the quantities, or to omit the unit. Angular momentum is the product of moment of inertia and angular velocity.

A Component 01 Section C extended-response question with a calculation. Markers reward the conservation principle, the moment-of-inertia change and the calculated angular velocities.

Award marks for: angular momentum is conserved in flight because, once airborne, no external torque acts (the only force, weight, acts through the centre of mass), so $H$ stays constant. Angular momentum is $H = I \times \omega$, so if moment of inertia decreases, angular velocity must increase, and vice versa. When the gymnast tucks (bringing the mass close to the axis), the moment of inertia falls and the spin speeds up; when they open out to land (mass far from the axis), the moment of inertia rises and the spin slows for a controlled landing. For example, with $H = 60$ kg m squared rad/s: in a layout with $I = 15$ kg m squared, $\omega = 60 \div 15 = 4$ rad/s; in a tuck with $I = 10$ kg m squared, $\omega = 60 \div 10 = 6$ rad/s, so the tuck spins faster.

A top answer states that angular momentum is conserved (no external torque), links the change in moment of inertia to the change in angular velocity, and supports it with a worked calculation.