91Ó°ÊÓ

Angular acceleration is the rate at which the angular velocity of an object changes with time. It's a crucial concept when dealing with rotational motions, especially when an object starts to rotate or speeds up. Think of it as the rotational equivalent of linear acceleration.

In the problem, the wheel starts from rest, implying its initial angular velocity (\( \omega_0 \) ) is 0. After 2 seconds, the wheel gains an angular velocity (\( \omega \) ) of \( 5 \mathrm{rad/s} \) . Using the formula \( \omega = \omega_0 + \alpha t \) , where \( \alpha \) is the angular acceleration, we substitute the known values to find \( \alpha = 2.5 \mathrm{rad/s^2} \) .

Angular acceleration is constant in this situation, meaning the wheel's angular velocity increases uniformly over time until the acceleration ceases. This type of motion is called uniform angular acceleration.

Angular velocity refers to how fast an object rotates or revolves around a point or axis, usually expressed in radians per second (\( \mathrm{rad/s} \) ). It's like the speedometer for rotational speed.

As the wheel starts from rest (initial angular velocity is zero), the constant angular acceleration increases its angular velocity steadily. So, at \( 2 \mathrm{s} \) , the wheel's angular velocity reaches \( 5 \mathrm{rad/s} \). This increment continues until \( 20 \mathrm{s} \), where the angular velocity becomes \( 50 \mathrm{rad/s} \).

• From \( 0 \rightarrow 20 \mathrm{s}, \) the wheel is under constant acceleration.
• From \( 20 \rightarrow 40 \mathrm{s}, \) this speed of \( 50 \mathrm{rad/s} \) remains constant since the angular acceleration stops.

The constant velocity from \( 20 \rightarrow 40 \mathrm{s} \) means the wheel covers equal angular displacement for each second.

Angular displacement is the measure of the angle through which an object moves on a circular path. It's a vital part of understanding rotational motion since it tells you how much the wheel turns.

To find the angular displacement of the wheel, we evaluate two separate intervals:

• From \( 0 \rightarrow 20 \mathrm{s} \): Using\( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \), where \( \omega_0 \) is \( 0, \alpha \) is \( 2.5 \mathrm{rad/s^2} \),andtis \( 20\mathrm{s}\),we calculate \( \theta = 500 \mathrm{rad} \).
• From \( 20 \rightarrow 40 \mathrm{s} \): With no more acceleration, \( \theta = \omega \times \Delta t \), where \( \omega = 50 \mathrm{rad/s} \) and \( \Delta t = 20 \mathrm{s} \), giving \( \theta = 1000 \mathrm{rad} \).

The total angular displacement is the sum of these two, which equals \( 1500 \mathrm{rad} \). This value represents how much the wheel has rotated in total over the \( 40 \mathrm{s} \) interval.