91Ó°ÊÓ

Moment of inertia, often denoted as \(I\), is a measure of how much torque is needed for a certain angular acceleration about a rotational axis. This concept is similar to mass in linear motion, as it quantifies the resistance to rotational movement.

For different shapes, the moment of inertia is calculated differently based on their geometry. In the case of a solid sphere, the formula is derived as \(I = \frac{2}{5}mr^2\). This means you multiply the mass \(m\) of the sphere by the square of its radius \(r\), and then multiply by \(\frac{2}{5}\).

• For our sphere: mass \(m = 75 \text{ kg}\), radius \(r = 0.10 \text{ m}\).
• Plugging values gives: \(I = \frac{2}{5} \times 75 \times 0.01 = 0.3 \text{ kg} \cdot \text{m}^2\).

This indicates it would take a significant torque to accelerate this sphere around its axis compared to a less massive or smaller object having a lower moment of inertia.

The torsional constant, represented by \(\kappa\), is a measure of the stiffness of a wire or filament when it is twisted. It describes how resistant the medium is to being twisted and how much torque will result from a unit twist angle.

For any object acting as a torsional pendulum, you can find the torsional constant by knowing the torque and the angle of twist. The formula to calculate it is \(\tau = \kappa \theta\), which can be rearranged as \(\kappa = \frac{\tau}{\theta}\). This shows us that torsional constant is directly proportional to the torque applied and inversely proportional to the twist angle.

• In our problem, \(\tau = 0.35 \text{ N} \cdot \text{m}\) and \(\theta = 0.85 \text{ rad}\).
• So, \(\kappa = \frac{0.35}{0.85} \approx 0.4118 \text{ N} \cdot \text{m/rad}\).

This constant helps in understanding how the wire will respond to twisting forces and is essential for calculating oscillation periods.

Torque, often symbolized as \(\tau\), is a measure of the rotational force applied to an object. It's similar to linear force but for rotational movement, essential in causing an object to rotate around an axis.

Imagine opening a door; you apply force at the knob, creating a torque that turns the door around its hinge. The further from the hinge (pivot point) you apply the force, the greater the torque, illustrating how force and distance from the pivot affect rotation.

• In our scenario, a torque of \(0.35 \text{ N} \cdot \text{m}\) is needed to rotate a sphere.
• This force, when applied, results in a specific amount of rotation due to the moment of inertia and torsional constant of the setup.

The relationship between torque and these factors can predict the oscillation behavior of the torsional pendulum once released from the twisted position.