91Ó°ÊÓ

Rotational Inertia, also known as the moment of inertia, measures how difficult it is to change the rotational speed of an object. It depends on the distribution of mass around the rotation axis.
For example, in our exercise, we have two objects: the motor rotor and the space probe. The motor rotor has a rotational inertia of \(I_{m} = 2.0 \times 10^{-3} \; \text{kg} \cdot \text{m}^{2}\), which is quite small. This suggests that less effort is required to change its rotational speed compared to the space probe.
The space probe, on the other hand, has a larger rotational inertia of \(I_{p} = 12 \; \text{kg} \cdot \text{m}^{2}\). This means it is significantly harder to adjust its rotational speed.

• The higher the rotational inertia, the more torque (which you can think of as rotational force) is needed to achieve the same angular acceleration.
• This is much like how it's easier to push a light toy car than a heavy truck.

The overall idea is that greater rotational inertia signifies that an object is resistant to changes in its angular state.

The principle of conservation of angular momentum states that if no external torque acts on a system, the total angular momentum remains constant. This concept is pivotal in our scenario where the space probe and the motor rotor function as a closed system.
According to the conservation rule, the change in angular momentum of the rotor \((\Delta L_{m})\) must match, in magnitude but opposite in direction, the change in angular momentum of the probe \((\Delta L_{p})\):\[ I_{m} \Delta \omega_{m} = -I_{p} \Delta \omega_{p} \]

• This means if the probe rotates in one direction, the rotor will rotate equivalently in the opposite direction to conserve angular momentum.
• It’s like the balancing act of a seesaw; if one side goes up, the other must go down to conserve balance.

In this exercise, it ensures that the actions of the rotor and the probe counterbalance each other perfectly.

Angular Displacement refers to the angle through which an object moves on a circular path. It shows us how far the object has turned or rotated from its initial position.
In our example, we need the space probe to turn by \(30^{\circ}\). However, we usually need to work in radians for calculations in physics. Thus, we convert:\[30^{\circ} = \frac{30 \pi}{180} = \frac{\pi}{6} \; \text{radians}\]
This conversion is essential for inserting the angular displacement into our equations.

• Degrees are often easier to visualize, but radians allow for more straightforward calculation in rotational kinematics.
• A complete circle is \(360^{\circ}\) or \(2\pi\) radians.

Expressing the needed rotation in radians provides clarity and a consistent base for further computation.

To understand how many times the rotor must spin, we calculate the number of revolutions it must make, given the angular distance it covers.
Since we know the angular displacement from the conservation equation, we connect it with revolutions.
We find the rotor’s angular displacement \(\theta_{m}\) is \(-1000\pi\) radians. To convert this into revolutions, we use the relationship:\[\text{Revolutions} = \frac{-1000\pi}{2\pi} = -500\]

• A negative sign in our solution signifies direction, not a reduction in revolution count. Therefore, the rotor requires 500 complete revolutions.
• This result indicates that effectively driving the probe to turn \(30^{\circ}\) involves extensive rotor motion, highlighting the coupling of angular actions between different components.

This key calculation connects the rotor’s motion in tangibly understandable terms of complete rotations or cycles.