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

To determine the angular speed ω in position 1, we can use the relationship derived from angular momentum conservation: $I_1 \cdot \omega_1 = I_2 \cdot \omega_2$.

Given that:

• Moment of inertia in position 1, $I_1 = 13.5 \, kg \, m^2$
• Moment of inertia in position 2, $I_2 = 4.1 \, kg \, m^2$
• Angular speed in position 2, $ext{ω}_2 = 14.2 \, rad/s$,

We can rearrange and solve for ω: $\omega_1 = \frac{I_2 \cdot \omega_2}{I_1} = \frac{4.1 \, kg \, m^2 \cdot 14.2 \, rad/s}{13.5 \, kg \, m^2} \approx 4.3 \, rad/s.$

Thus, the angular speed of the gymnast in position 1 is approximately 4.3 rad/s.

To find the number of complete rotations, we first calculate the angular distance covered in time (t) using the angular speed in position 2:

We use the formula: $\theta = \omega \cdot t$

where:

• Angular speed, $\omega = 14.2 \, rad/s$,
• Time, $t = 1.2 \, s$.

Calculating: $\theta = 14.2 \, rad/s \cdot 1.2 \, s = 17.04 \, rad.$

To convert radians to complete rotations, divide by $2\pi$: $\text{Number of rotations} = \frac{\theta}{2\pi} = \frac{17.04 \, rad}{2\pi} \approx 2.71 \text{ rotations}.$

Hence, the gymnast completes 2 complete rotations.

1. Increase Initial Angular Speed: The gymnast can increase the initial angular speed before entering the tuck position. This can be achieved by gaining more height and angular momentum during the swing on the bar, leading to a greater rate of rotation.

2. Optimize Body Posture: The gymnast can also ensure a more effective tuck position by bringing their arms and knees closer to their body, which further reduces their moment of inertia. This helps in accelerating the rotation as they maintain a compacted form during the tuck.