Figure 3 shows the basic principle of operation of a hand-operated salad spinner used to dry washed salads - AQA - A-Level Physics - Question 2 - 2017 - Paper 6

Since gear C rotates four times for every one revolution of gear B, the torque relationship can be examined. The torque is related by the gear ratio. More specifically, if we denote the torque on gear B as $T_B$ and that on gear C as $T_C$, the maximum theoretical torque relationship can be expressed as:

$T_C = T_B \times 4$

However, due to the conservation of energy, the power cannot increase, which implies that:

$T_C \leq T_B$

Thus, it is not possible for $T_C$ to be greater than $T_B$.

To find the moment of inertia $I$ of the basket, we can use the relationship between torque ($\tau$), angular acceleration ($\alpha$), and moment of inertia:

$\tau = I \alpha$

We can calculate angular acceleration using:

$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{76 \text{ rad s}^{-1}}{2.1 \, ext{s}} \approx 36.19 \, ext{rad s}^{-2}$

Now, we can use the torque for gear C (0.04 N m) to determine the moment of inertia:

$0.04 = I \times 36.19$

Solving for $I$ gives:

$I \approx \frac{0.04}{36.19} \approx 0.0011 \, ext{kg m}^2$

Angular impulse can be defined as the change in angular momentum ($L$) of an object, expressed as:

$\Delta L = \tau \Delta t$

To stop the basket quickly, a greater force leads to a larger torque, which results in a greater angular deceleration. This rapid deceleration creates a large angular impulse. Specifically,

$\Delta L = I \Delta \omega$

where $\Delta \omega$ represents the change in angular speed. When the handle is pulled quickly to stop the basket, the gear exerting a high force is necessary to produce sufficient torque within a small time frame. This ensures a quick halt to the loaded basket, preventing overspinning.