Question

current attempt in progress
which one of the following equations is only valid if and when the angular measure is expressed in radians?
$\alpha = \frac{\delta \omega}{\delta t}$
$\omega = \frac{\delta \theta}{\delta t}$
$\theta = \frac{1}{2} \alpha t^{2}+\omega_{0} t$
$\omega = \frac{v t}{r}$
$\omega^{2}=\omega_{0}^{2}+2 \alpha \theta$

Explanation:

Step1: Recall the nature of angular measure in formulas

• For $\alpha=\frac{\Delta\omega}{\Delta t}$: Angular acceleration formula. $\alpha$ (angular acceleration), $\omega$ (angular velocity) and $t$ (time) are defined in a way that $\alpha=\frac{\Delta\omega}{\Delta t}$ is a general - rate - of - change formula. It does not depend on the unit of angular measure (radians or degrees) for its validity.
• For $\omega=\frac{\Delta\theta}{\Delta t}$: Angular velocity formula. Similar to the linear velocity formula $v = \frac{\Delta x}{\Delta t}$, it is a rate - of - change formula. The unit of $\theta$ (whether radians or degrees) does not affect the fundamental rate - of - change relationship in the formula $\omega=\frac{\Delta\theta}{\Delta t}$ in terms of its mathematical structure (as a ratio of changes).
• For $\theta=\frac{1}{2}\alpha t^{2}+\omega_{0}t$: This is the angular - displacement formula derived from the integration of angular - velocity and angular - acceleration equations (assuming constant angular acceleration). The derivation of the kinematic equations for rotational motion (such as $\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}$, $\omega=\omega_{0}+\alpha t$, $\omega^{2}=\omega_{0}^{2} + 2\alpha\theta$) is based on the fact that when we use calculus (integration and differentiation) for rotational motion, the angular measure must be in radians. For example, when we start from the definition of angular velocity $\omega=\frac{d\theta}{dt}$ and angular acceleration $\alpha=\frac{d\omega}{dt}$, and integrate $\omega=\omega_{0}+\alpha t$ to get $\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}$, the small - angle approximations and the calculus operations (like $\int\omega d t$) are valid only when $\theta$ is in radians.
• For $\omega=\frac{vT}{r}$: We know that $v = r\omega$ and $T=\frac{2\pi}{\omega}$. Then $\omega=\frac{vT}{r}=\frac{v\times\frac{2\pi}{\omega}}{r}$, or $\omega^{2}=\frac{2\pi v}{r}$. Also, from $s = r\theta$ (where $s$ is arc - length, $r$ is radius and $\theta$ is in radians), $v=\frac{s}{t}=\frac{r\theta}{t}=r\omega$ (when $\theta$ is in radians). But the formula $\omega=\frac{vT}{r}$ can be re - arranged and verified without relying on the radian - only property in its basic form (using the relationships between linear and rotational quantities and the period $T$).
• For $\omega^{2}=\omega_{0}^{2}+2\alpha\theta$: This is also a kinematic equation for rotational motion derived from calculus (by eliminating the time variable $t$ from $\omega=\omega_{0}+\alpha t$ and $\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}$). The validity of the calculus - based derivations of rotational kinematic equations (including $\theta=\frac{1}{2}\alpha t^{2}+\omega_{0}t$) requires that the angular measure $\theta$ is in radians.

Answer:

$\theta=\frac{1}{2}\alpha t^{2}+\omega_{0}t$