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 relative to the axis of rotation. The screw, punch, and arms have their mass concentrated closer to the axis compared to the steel balls, which are located further out. Therefore, the distance from the axis contributes significantly to the moment of inertia.

In mathematical terms, since moment of inertia is defined as:

$I = \sum m r^2$

where (m) is the mass and (r) is the distance from the axis. Since the steel balls have a larger (r), their moment of inertia is much greater.

To find the angular deceleration ((\alpha)), we use the equation:

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

Where:

• (\Delta \omega = 0 - 2.9 \text{ rev s}^{-1} = -2.9 \text{ rev s}^{-1} \approx -18.2 \text{ rad s}^{-1}$$
• (\Delta t = 89 \text{ ms} = 0.089 ext{ s}$$

Substituting in:

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

For angular displacement ((\theta)), we can use the kinematic equation:

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

Where:

• (\omega_i = 2.9 \text{ rev s}^{-1} \approx 18.2 \text{ rad s}^{-1}$$
• (t = 0.089 ext{ s}$$
• (\alpha = -204.5 \text{ rad s}^{-2}$$

Substituting in:

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

Calculating:

$\theta \approx 1.62 - 0.81 \approx 0.81 ext{ rad}$

From the formula for rotational kinetic energy:

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

When increasing (y) by 15%, the moment of inertia primarily remains the same, but the distance from the axis increases slightly, leading to a marginal increase in energy. On the other hand, increasing (R) directly affects the moment of inertia as:

$I \propto R^2$

Thus, increasing (R) by 15% will produce a larger increase in stored energy overall as it affects the inertia more significantly due to the square relationship.

The SI unit for angular impulse is represented as: N s. Therefore, tick the box beside N s.