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 find the linear velocity $I$ after the yo-yo has fallen 0.5 m, we apply conservation of energy principles. The initial potential energy (PE) of the yo-yo when it is held stationary is converted into both translational and rotational kinetic energy (KE) as it falls.

The potential energy lost is given by:

$PE = mgh$

where:

• $m = 9.2 \times 10^2$ kg (mass of the yo-yo)
• $g = 9.81$ m/s² (acceleration due to gravity)
• $h = 0.5$ m (fall height)

Calculating this gives:

$PE = (9.2 \times 10^2) \times 9.81 \times 0.5 = 4510.41$ J

The kinetic energy gained can be represented as the sum of translational and rotational kinetic energies:

$KE_{total} = KE_{translational} + KE_{rotational} = \frac{1}{2} mv^2 + \frac{1}{2} I \beta^2$

Substituting $v = r \beta$ into the energy equation:

$KE_{total} = \frac{1}{2} mv^2 + \frac{1}{2} \cdot 8.6 \times 10^5 \cdot \frac{v^2}{r^2}$

Equating initial potential energy to the total kinetic energy:

$4510.41 = \frac{1}{2} (9.2 \times 10^2) v^2 + \frac{1}{2} (8.6 \times 10^5) \frac{v^2}{(5 \times 10^{-3})^2}$

Now, let's solve for $v$. After algebraic manipulations, the linear velocity $I$ at which the yo-yo is moving after it has fallen 0.5 m is calculated to be approximately 0.51 m/s.