91Ó°ÊÓ

When a rotating object slows down, we describe this change in motion with angular deceleration. This is denoted by the symbol \( \alpha \), and is measured in radians per second squared \( \text{rad/s}^2 \). In this context, it represents how quickly the angular speed of the object decreases over time.
For a wheel initially rotating at \( 15 \, \text{rad/s} \) and halting after \( 5 \) seconds, we use the formula \( \alpha = \frac{\omega_f - \omega_i}{t} \). Here, \( \omega_i \) is the initial angular speed, \( \omega_f \) is the final angular speed (zero when stopped), and \( t \) is the time.

• Initial speed \( \omega_i = 15 \, \text{rad/s} \)
• Final speed \( \omega_f = 0 \, \text{rad/s} \)
• Time \( t = 5 \, \text{s} \)

Substituting, \( \alpha = \frac{0 - 15}{5} = -3 \, \text{rad/s}^2 \), indicating the wheel is decreasing its speed by \( 3 \, \text{rad/s} \) every second. The negative sign shows it's decelerating.

Kinetic energy in rotational motion depends on how fast an object spins and its rotational inertia, also known as the moment of inertia \( I \). For rotational motion, kinetic energy \( KE \) can be calculated using the formula \( KE = \frac{1}{2} I \omega^2 \), where \( \omega \) is the angular speed.
Initially, with an angular speed of \( 15 \, \text{rad/s} \), the wheel's kinetic energy is:

• \( KE_i = \frac{1}{2} \times 0.50 \times 15^2 = 56.25 \, \text{J} \)

After \( 3 \) seconds, the angular speed drops to \( 6 \, \text{rad/s} \), reducing the energy:

• \( KE_f = \frac{1}{2} \times 0.50 \times 6^2 = 9.0 \, \text{J} \)

The change in kinetic energy shows the amount of energy required to slow down the wheel, crucial for understanding how much work is done by the external force.

Work is a measure of energy transfer. In rotational dynamics, the work done on an object is calculated as the change in its kinetic energy. This means the effort applied causes a shift in the wheel’s energy state. Initially, the wheel had a kinetic energy of \( 56.25 \, \text{J} \), and after slowing to \( t = 3 \, \text{s} \), its kinetic energy decreases to \( 9 \, \text{J} \).
This decrease equates to work done, denoted by \( W \), and calculated as:

• \( W = KE_i - KE_f = 56.25 \, \text{J} - 9.0 \, \text{J} = 47.25 \, \text{J} \)

Thus, \( 47.25 \, \text{J} \) of work was done by the man to bring the wheel to a reduced state of motion at \( 3 \, \text{s} \), and ultimately \( 56.25 \, \text{J} \) to stop it completely. This demonstrates how much energy needs to be removed to cease the rotation.

Power is the rate at which work is done or energy is transferred. When considering average power over a period of time \( t \), it’s determined by dividing total work done \( W \) by the time interval \( t \).
The concept is crucial when you need to understand how efficiently work is performed within a given time frame.

• Total work \( W_{total} = 56.25 \, \text{J} \)
• Time \( t = 5 \, \text{s} \)
• Average power \( P = \frac{W_{total}}{t} = \frac{56.25}{5} = 11.25 \, \text{W} \)

This implies, on average, \( 11.25 \, \text{W} \) of power was required to bring the rotating wheel to a standstill throughout the \( 5 \, \text{s} \) interval. The concept of power helps us understand how much energy "intensity" an operation, like reducing rotation, necessitates.