91Ó°ÊÓ

Angular velocity is an important concept in rotational motion. It is the rate at which an object rotates or revolves around an axis. In mathematical terms, it is the derivative of the angular position with respect to time. For our thin-walled hollow sphere, the given function for angular displacement is \( \theta(t) = A t^2 + B t^4 \).The angular velocity \( \omega(t) \) is calculated by taking the derivative of \( \theta(t) \) concerning time:

• \( \omega(t) = \frac{d\theta}{dt} = 2At + 4Bt^3 \)

Substituting the values of \( A = 1.50 \) and \( B = 1.10 \), we find:

• \( \omega(t) = 3.00t + 4.40t^3 \)

At \( t = 3.00 \) seconds, the angular velocity becomes:

• \( \omega(3.00) = 3.00(3.00) + 4.40(3.00)^3 \)

Calculate this value to get the specific angular velocity at 3 seconds. This measure tells us how fast the sphere is spinning at that particular moment.

Net torque is a measure of how much a force acting on an object causes that object to rotate. It is the rotational equivalent of linear force. The torque \( \tau \) is calculated using the formula:

• \( \tau = I \cdot \alpha(t) \)

where \( I \) is the moment of inertia, and \( \alpha(t) \) is the angular acceleration.
Angular acceleration \( \alpha(t) \) is derived by taking the derivative of the angular velocity with respect to time:

• \( \alpha(t) = \frac{d\omega}{dt} = 3.00 + 13.20t^2 \)

At \( t = 3.00 \), the angular acceleration is:

• \( \alpha(3.00) = 3.00 + 13.20(3.00)^2 \)

Using this value of \( \alpha(3.00) \) alongside the moment of inertia \( I \), calculated for the sphere, you can then determine the net torque. The torque affects how efficiently the sphere continues to rotate as time progresses.

Moment of inertia is akin to mass in linear motion but applies to rotational motion. It represents the resistance of a body to change its state of motion about an axis. For a hollow, thin-walled sphere, the formula used to calculate the moment of inertia \( I \) is:

• \( I = \frac{2}{3}mr^2 \)

where \( m \) is the mass and \( r \) is the radius.
In our problem, the sphere has a mass \( m = 12.0 \) kg and a diameter of 48.0 cm, which translates to a radius \( r = 0.24 \) m:

• \( I = \frac{2}{3} \times 12.0 \times (0.24)^2 \)

This gives the specific moment of inertia for the sphere, which can then be used to compute other properties such as angular momentum and torque. Understanding the moment of inertia helps us grasp why different objects, even with the same mass, rotate differently.