Figure 2 shows a yo-yo made of two discs separated by a cylindrical axle - AQA - A-Level Physics - Question 2 - 2021 - Paper 6

To calculate the linear velocity $I$ of the yo-yo after falling 0.50 m, we start by considering the energy transfers involved in the fall.

1. Initial Potential Energy (PE): The potential energy of the yo-yo at the height before falling is given by: $PE = mgh$ where:

   • $m = 9.2 \times 10^{-2} \text{ kg}$ (mass of the yo-yo)
   • $g = 9.81 \text{ m/s}^2$ (acceleration due to gravity)
   • $h = 0.50 \text{ m}$ (height fallen)

   Substituting the values: $PE = (9.2 \times 10^{-2}) \times 9.81 \times 0.50 = 0.451 \text{ J}$

2. Final Kinetic Energy (KE): As the yo-yo falls, this potential energy converts into kinetic energy, which is the sum of translational and rotational kinetic energy: $KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$ Here, $v$ is the linear velocity, $I$ is the moment of inertia, and $\omega$ is the angular velocity. The relationship between $v$ and $\omega$ is given by $v = r\omega$, where $r = 5.0 \times 10^{-3} \text{ m}$. Thus: $KE = \frac{1}{2} mv^2 + \frac{1}{2} I \left(\frac{v}{r}\right)^2$

3. Substituting Known Values: The moment of inertia $I = 8.6 \times 10^{-5} \text{ kg m}^2$ and we have the mass $m = 9.2 \times 10^{-2} \text{ kg}$ and $r = 5.0 \times 10^{-3} \text{ m}$. By setting the potential energy equal to the kinetic energy: $0.451 = \frac{1}{2} (9.2 \times 10^{-2}) v^2 + \frac{1}{2} (8.6 \times 10^{-5}) \left(\frac{v}{5.0 \times 10^{-3}}\right)^2$

4. Solving for $v$: Rearranging and solving this equation will give the value of $v$ which can be used to determine $I$. After performing these calculations, we find: $v = 0.51 \text{ m/s}$