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 analyzed. If the input torque on gear B is $T_B$, then the torque on gear C ($T_C$) must satisfy:

$T_C = T_B \times 4$

However, this would imply that the mechanical advantage cannot exceed four times without additional energy input. Thus, the torque on gear C cannot be greater than that on gear B under normal circumstances due to the conservation of power, implying that the torque on gear C must be equal to or less than the input torque when accounted for efficiency losses.

The moment of inertia ($I$) can be found using the rotational analog of Newton's second law:

$\tau = I \alpha$

where:

• $au = 0.054 \, \text{N m}$
• rac{\Delta \omega}{\Delta t} = \alpha = \frac{\Delta \omega}{t} = \frac{76 \, \text{rad s}^{-1}}{2.1 \, \text{s}} \approx 36.19 \, \text{rad s}^{-2}

Rearranging for $I$ gives:

$I = \frac{\tau}{\alpha} = \frac{0.054}{36.19} \approx 0.00149 \, \text{kg m}^2$

When a sudden need occurs to stop the loaded basket, angular impulse comes into play. Angular impulse is defined as:

$J = \Delta L = I \Delta \omega$

where $J$ is the angular impulse, $I$ is the moment of inertia, and $\\Delta \omega$ is the change in angular velocity. If the basket is moving with significant angular velocity, a large change in momentum needs to occur to stop it quickly. This requires a great torque applied to the gear teeth, resulting in high forces which can stress and damage the plastic gears if they are not designed to handle such sudden loads.