91Ó°ÊÓ

Angular deceleration refers to the rate at which an object's rotational speed decreases over time. It is similar to linear deceleration but applies to rotating objects. In our problem, the wheel experiences a retarding or slowing force due to friction.
The formula connecting torque (\( \tau \)), moment of inertia (\( I \)), and angular deceleration (\( \alpha \)) is:

• \( \tau = I \cdot \alpha \)

Here, torque is the measure of the rotational force acting to decelerate the wheel. Friction provides a negative torque, which reduces the wheel's rotation speed until it comes to rest. The negative sign indicates that the effect is to slow down the motion.

• In the exercise, a torque of \(-0.12 \, \mathrm{N} \cdot \mathrm{m}\) causes an angular deceleration \( \alpha = -0.75 \, \mathrm{rad/s}^2 \).

The time needed for the wheel to stop can be found using the rotational kinematic equation:

• \( \omega_f = \omega_i + \alpha t \)

where \( \omega_f \) is the final angular velocity (0 in our case), \( \omega_i \) is the initial angular velocity, and \( t \) is the time.

Moment of inertia is crucial in understanding rotational motions. It is the rotational analog of mass in linear motion and defines how the mass is distributed with respect to the axis of rotation. The greater the moment of inertia, the harder it is to change the rotational speed of the object.
The moment of inertia for a rigid body is calculated using the formula:

• \( I = m \cdot k^2 \)

where \( m \) is the mass of the object and \( k \) is the radius of gyration (a measure of the distribution of the object's mass around the axis of rotation).

In the given exercise:

• The mass \( m \) of the wheel is 4.0 kg.
• The radius of gyration \( k \) is 0.2 m.
• This results in a moment of inertia \( I = 0.16 \, \text{kg} \cdot \text{m}^2 \).

The moment of inertia thus calculates how resistant the wheel is to changes in its angular motion, which in this problem is necessary to determine the angular deceleration.

Rotational kinetic energy is the energy that an object possesses due to its rotation. It can be thought of as the rotational equivalent of linear kinetic energy. Knowing this helps you understand how energy conversions occur in rotating systems.
The rotational kinetic energy \( K \) of a rotating body is given by:

• \( K = \frac{1}{2} I \omega^2 \)

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

In the scenario of our exercise:

• Initial \( \omega \) is 37.7 rad/s.
• The moment of inertia \( I \) is 0.16 kgm².
• Thus, the initial rotational kinetic energy is \( \frac{1}{2} \times 0.16 \times (37.7)^2 \).

As the wheel slows down and comes to a stop, its rotational kinetic energy is transferred due to friction into other forms like thermal energy, eventually dropping to zero when the wheel is at rest.