Rotational kinematics

Rotational motion is the angular twin of motion along a line. Angular position $\theta$ says how far an object has turned, angular velocity $\omega$ says how fast it is turning, and angular acceleration $\alpha$ says how that turning rate is changing.

The three are linked by calculus, just like position, velocity, and acceleration: $\omega = d\theta/dt$ and $\alpha = d\omega/dt$. Run the chain the other way by integrating: $\omega$ is the integral of $\alpha$, and $\theta$ is the integral of $\omega$.

For students arriving from algebra-based Physics 1, the constant-$\alpha$ equations will look familiar. The Physics C addition is the calculus: when $\alpha$ changes with time you differentiate and integrate, and you keep the constants of integration, $\omega_0$ and $\theta_0$, every time.

Measure angles in radians. Angular position $\theta$ is the angle from a chosen reference, positive in one direction (take counterclockwise as positive) and negative in the other.

Angular velocity is the rate of change of angular position, $\omega = \dfrac{d\theta}{dt}$, in radians per second. Angular acceleration is the rate of change of angular velocity, $\alpha = \dfrac{d\omega}{dt}$, in radians per second squared. A positive $\omega$ means the angle is growing; a positive $\alpha$ means $\omega$ is growing.

Going backward, the definitions become integrals. The angular velocity comes from the angular acceleration, and the angle from the angular velocity: $\omega(t) = \omega_0 + \int_0^t \alpha\,dt'$ and $\theta(t) = \theta_0 + \int_0^t \omega\,dt'$. The starting values $\omega_0$ and $\theta_0$ are part of the answer, not optional extras.

Angular velocity is the derivative of angular position, and angular acceleration is the derivative of angular velocity. Reverse the chain by integrating, and carry $\omega_0$ and $\theta_0$.

When the angular acceleration is constant, integrating gives three familiar equations, the angular versions of the straight-line kinematics formulas: $\omega = \omega_0 + \alpha t$, then $\theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2$, and $\omega^2 = \omega_0^2 + 2\alpha\,\Delta\theta$.

Each one is derived on the assumption that $\alpha$ does not change. If $\alpha$ is constant, they are the fastest route to $\omega$ and $\theta$. If $\alpha$ varies with time, they do not apply, and reaching for them anyway is the most common mistake in this topic.

A common slip is to read a coefficient as if it were the constant $\alpha$. If a problem states $\alpha = 4t$, the number $4$ is not the angular acceleration; $\alpha$ is changing every instant, and $\omega = \omega_0 + \alpha t$ is off the table.

The constant-$\alpha$ equations apply only when $\alpha$ is constant. Check that first; if $\alpha$ depends on time, integrate instead.

If $\alpha$ is a function of time, find $\omega$ and $\theta$ by integrating, one step at a time. Integrate the angular acceleration to get the angular velocity, then integrate the angular velocity to get the angle: $\omega(t) = \omega_0 + \int_0^t \alpha(t')\,dt'$ and $\theta(t) = \theta_0 + \int_0^t \omega(t')\,dt'$.

Worked example. Suppose $\alpha(t) = 6t$ rad/s$^2$, with $\omega_0 = 2$ rad/s and $\theta_0 = 5$ rad. Then $\omega = 2 + \int_0^t 6t'\,dt' = 2 + 3t^2$, and $\theta = 5 + \int_0^t (2 + 3t'^2)\,dt' = 5 + 2t + t^3$. At $t = 2$ s, $\omega = 14$ rad/s and $\theta = 17$ rad.

The two added constants, $\omega_0$ and $\theta_0$, are the constants of integration. Each integral gives only the change in the quantity; the starting value is what turns that change into the actual function. Drop one and the answer is wrong by exactly that starting value.

For time-varying $\alpha$, integrate up the chain and add the initial value at each step. The integral gives the change; $\omega_0$ and $\theta_0$ supply the rest.

Whether a rotating object is speeding up or slowing down is not decided by the sign of $\alpha$ alone. It depends on how the sign of $\alpha$ compares with the sign of $\omega$.

The object speeds up when $\omega$ and $\alpha$ share a sign, and slows down when they have opposite signs. A negative $\alpha$ slows an object that is spinning the positive way, but it speeds up an object already spinning the negative way. A zero $\alpha$ means $\omega$ is unchanged, neither speeding up nor slowing.

When $\omega$ and $\alpha$ oppose, the object slows, can reach $\omega = 0$ for an instant, and then reverses. Past that turning point the two signs agree again, and the object speeds up in the new direction. This is the rotational version of a ball thrown upward slowing, stopping, and falling back.

Compare the signs of $\omega$ and $\alpha$. Same sign means speeding up; opposite signs mean slowing. The sign of $\alpha$ by itself decides nothing.

Angular kinematics fails in a handful of familiar ways—each pitfall below is tagged with the code the diagnostic watches for.