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Angular Kinematic Equations (Constant \( \alpha \))

When angular acceleration \( \alpha \) is constant, we can use a set of equations — known as the angular kinematic equations — to solve problems in rotational motion. These equations are direct analogs of the linear kinematics equations you're familiar with.

The Angular Kinematic Equations

\( \omega = \omega_0 + \alpha t \)

This relates angular velocity to time under constant angular acceleration.

\( \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \)

This gives angular displacement based on time, initial angular velocity, and angular acceleration.

\( \omega^2 = \omega_0^2 + 2 \alpha (\theta - \theta_0) \)

This equation connects angular displacement and angular velocity without explicitly involving time.

Analogous Linear Forms

These equations are directly analogous to their linear counterparts:

\( v = v_0 + a t \)
\( x = x_0 + v_0 t + \frac{1}{2} a t^2 \)
\( v^2 = v_0^2 + 2 a (x - x_0) \)

This close connection makes it easier to learn rotational motion by leveraging what you already know from linear motion.

Relating Rotational and Linear Quantities

Linear and rotational quantities are linked through the radius \( r \) of the circular path:

Linear displacement \( x \) is related to angular displacement \( \theta \) by \( x = r \theta \)
Linear velocity \( v \) is related to angular velocity \( \omega \) by \( v = r \omega \)
Linear acceleration \( a \) is related to angular acceleration \( \alpha \) by \( a = r \alpha \)

Example Problem: Rotational to Linear Conversion

A wheel of radius \( r = 0.5 \, \text{m} \) starts from rest and accelerates with a constant angular acceleration \( \alpha = 2 \, \text{rad/s}^2 \). Calculate:

The angular velocity after 4 seconds.
The angular displacement after 4 seconds.
The linear velocity and linear displacement of a point on the edge of the wheel after 4 seconds.

\(\omega = \omega_0 + \alpha t = 0 + 2 \times 4 = 8 \, \text{rad/s}\)

\(\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 = 0 + 0 + \frac{1}{2} \times 2 \times 16 = 16 \, \text{rad}\)

\(v = r \omega = 0.5 \times 8 = 4 \, \text{m/s}\)

\(x = r \theta = 0.5 \times 16 = 8 \, \text{m}\)

So after 4 seconds, the point on the edge of the wheel moves with a linear velocity of 4 m/s and has traveled 8 meters along its circular path.

Key Takeaway

By relating angular and linear quantities through the radius, you can solve rotational problems and directly find the corresponding linear motion of points on a rotating object.

Spinning Disk Simulation

Initial Angular Velocity \( \omega_0 \) (rad/s): 0

Angular Acceleration \( \alpha \) (rad/s²): 0