A gymnast dismounts from an exercise in which he swings on a high bar - AQA - A-Level Physics - Question 1 - 2021 - Paper 6

Using the relationship of angular momentum, we have: $I_1ω_1 = I_2ω_2$ Substituting the known values: $13.5ω_1 = 4.1 × 14.2$ Solving for ω_1 gives: $ω_1 = \frac{4.1 × 14.2}{13.5} \approx 1.04 \text{ rad/s}$

The gymnast stays in the tuck for 1.2 seconds, so we calculate the total rotations: First, find the angular speed in radians per second: $ω = 14.2 \text{ rad/s}$ Next, determine the total angle rotated: $\text{Total angle} = ω × t = 14.2 \times 1.2 = 17.04 \text{ radians}$ To convert this to complete rotations, divide by $2π$: $\text{Complete rotations} = \frac{17.04}{2π} \approx 2.71 \text{ rotations}$

1. Increase initial angular speed: By initiating the dismount with a higher angular speed before tucking, the gymnast maximizes rotational momentum, allowing for more rotations before landing.
2. Maintain a tighter tuck position: By drawing their knees closer to the chest in a tighter tuck, the moment of inertia decreases further, which would lead to an increase in angular speed, enabling the gymnast to complete more rotations.