The fly-press shown in Figure 1 is used by a jeweller to punch shapes out of a thin metal sheet - AQA - A-Level Physics - Question 1 - 2019 - Paper 6

The moment of inertia depends on how mass is distributed about the axis of rotation. The mass of the screw, punch, and arms is concentrated close to the axis of rotation, resulting in a smaller moment of inertia. In contrast, the steel balls are located further from the axis, which significantly increases their moment of inertia due to the $r^2$ term in the moment of inertia formula. Therefore, $I \propto r^2$, where r is the distance from the axis of rotation.

To find angular deceleration ($\alpha$):

Using the formula:

$\alpha = \frac{\Delta \omega}{\Delta t}$

where $\Delta \omega = 0 - 18.2 = -18.2 ext{ rad s}^{-1}$

and $\Delta t = 89 ext{ ms} = 0.089 ext{ s}$,

we get:

$\alpha = \frac{-18.2}{0.089} \approx -204.5 ext{ rad s}^{-2}$

For the angle turned ($\theta$):

Using:

$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$

Substituting gives:

$\theta = (18.2)(0.089) + \frac{1}{2}(-204.5)(0.089)^2 \approx 0.55 ext{ rad}$

The stored energy in a rotating object is given by:

$E = \frac{1}{2} I \omega^2$

If we increase $y$ by 15% without changing $R$, the distance to center decreases the moment of inertia less significantly than if we increase $R$ by 15%. The impact on energy is given by the ratios:

• With $y$ increased: $I' = I(1 + 0.15)$
• With $R$ increased: $I' = I(1.15)^2$

Since $I' = I(1.15)^2$ will lead to a greater moment of inertia increase, thus the increase in stored energy will be greater with the increase in $R$.

The correct SI unit for angular impulse is N m s. Therefore, tick the box next to:

• N m s