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 the distribution of mass relative to the axis of rotation. The screw, punch, and arms have their mass concentrated closer to the axis of rotation, leading to a lower moment of inertia. In contrast, the steel balls are further away from this axis, resulting in a higher moment of inertia. Thus, the more mass is distributed away from the axis, the larger the moment of inertia.

We can use the kinematic equation for angular motion:

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

Where:

• (\Delta \omega = 0 - 2.9{ rev s}^{-1} = -2.9{ rev s}^{-1})
• (\Delta t = 89 ms = 0.089 s)

Converting (\Delta \omega) to radians:

$\Delta \omega = -2.9 \times 2\pi = -18.224 \text{ rad s}^{-1}$

Now substituting into the equation:

$\alpha = \frac{-18.224}{0.089} \approx -204.77 \, \text{rad s}^{-2}$

Using the formula for angular displacement:

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

Substituting the known values:

• Initial angular velocity, (\omega = 18.224 , \text{rad s}^{-1})
• Time, (t = 0.089 , s)
• Angular acceleration, (\alpha = -204.77 , \text{rad s}^{-2})

We have:

$\theta = 18.224 \times 0.089 + \frac{1}{2} \times (-204.77) \times (0.089)^2$

Calculating that gives:

$\theta \approx 1.623 - 0.8045 \approx 0.8185 \, ext{rad}$

The stored energy (E) in the system is related to the moment of inertia and angular velocity:

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

If we increase (y) by 15% without changing (R), it will result in a decrease of torque producing less energy than increasing (R) by 15% without changing (y). This is because the effective distance from the axis increases significantly when increasing the radius, thus leading to a greater increase in stored energy. Therefore, increasing (R) by 15% will produce the greater increase in stored energy.

The SI unit for angular impulse is N s. Thus, the correct option is:

• N s