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question 20
which equation is valid only when the angular measure is expressed in radians?
$\omega=\frac{\delta\theta}{\delta t}$
$\omega^{2}=\omega_{0}^{2}+2\alpha\theta$
$\alpha=\frac{\delta\omega}{\delta t}$
$\omega\frac{v_{t}}{r}$
$\theta=\omega_{o}t+\frac{1}{2}\alpha t^{2}$

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# Brief Explanations:
To determine which equation is valid only in radians, we analyze each:
- $ \omega = \frac{\Delta\theta}{\Delta t} $: Defines angular velocity, holds for any angular unit (since it’s a ratio of change in angle to time).
- $ \omega^2 = \omega_0^2 + 2\alpha\theta $: Derived from rotational kinematics, but the relationship relies on the **radian** for the calculus-based derivation (where $ s = r\theta $ holds only in radians, and this equation is analogous to linear $ v^2 = v_0^2 + 2ax $, which requires consistent units). Wait, no—actually, the key is the equation $ \omega = \frac{v_T}{r} $ (corrected from the typo $ \omega \frac{v_T}{r} $ to $ \omega = \frac{v_T}{r} $): The tangential speed $ v_T = r\omega $ holds **only when $ \theta $ is in radians** (because $ s = r\theta $ (arc length) is valid in radians, leading to $ v_T = \frac{ds}{dt} = r\frac{d\theta}{dt} = r\omega $). Wait, re-examining the options: The last valid-looking one is $ \theta = \omega_o t + \frac{1}{2}\alpha t^2 $, but no—let’s re-express.

Wait, the correct logic: Equations like $ v_T = r\omega $ (or $ \omega = \frac{v_T}{r} $) rely on $ s = r\theta $, which is true only when $ \theta $ is in radians (since $ s $ (arc length) and $ r $ (radius) have the same units, so $ \theta $ must be unitless, which radians are). However, among the options, the equation $ \omega = \frac{v_T}{r} $ (written as $ \omega \frac{v_T}{r} $ in the question, likely a typo for $ \omega = \frac{v_T}{r} $) is valid **only in radians** because $ v_T = r\omega $ comes from $ s = r\theta $ (arc length), where $ \theta $ must be in radians for $ s $ and $ r $ to have consistent units (radians are unitless, so $ s $ (length) = $ r $ (length) × $ \theta $ (unitless) works).

Other equations:
- $ \alpha = \frac{\Delta\omega}{\Delta t} $: Defines angular acceleration, unit-independent (ratio of angular velocity change to time).
- $ \theta = \omega_o t + \frac{1}{2}\alpha t^2 $: Rotational kinematic equation, analogous to linear $ x = v_0 t + \frac{1}{2}at^2 $; the derivation assumes calculus (where $ \omega = \frac{d\theta}{dt} $, $ \alpha = \frac{d\omega}{dt} $), but the equation itself holds for any angular unit if the units of $ \omega $, $ \alpha $, and $ t $ are consistent.

The key equation valid **only in radians** is $ \omega = \frac{v_T}{r} $ (or the kinematic equations that rely on $ s = r\theta $ for their derivation, but among the options, the correct one is $ \omega = \frac{v_T}{r} $ (the fourth option, likely a typo in the question as $ \omega \frac{v_T}{r} $ should be $ \omega = \frac{v_T}{r} $). Alternatively, the rotational kinematic equation $ \theta = \omega_o t + \frac{1}{2}\alpha t^2 $ is valid for any angular unit if the units of $ \omega $, $ \alpha $, and $ t $ are consistent. Wait, no—let’s check standard physics: The relationship $ v_T = r\omega $ (tangential speed = radius × angular velocity) is valid **only when $ \theta $ is in radians** because $ s = r\theta $ (arc length) requires $ \theta $ in radians (so that $ s $ and $ r $ have the same units, making $ \theta $ unitless). Thus, the equation $ \omega = \frac{v_T}{r} $ (the fourth option, despite the typo) is valid only in radians.

Wait, the options given: Let’s re-express the fourth option as $ \omega = \frac{v_T}{r} $ (correcting the typo). Among the options, this is the one that depends on radians because it comes from $ v_T = \frac{ds}{dt} = r\frac{d\theta}{dt} = r\omega $, which requires $ \theta $ in radians (since $ ds = r d\theta $ holds for radians).

# Answer:
$ \boldsymbol{\omega = \frac{v_T}{r}} $ (corresponding to the fourth option, likely written as $ \omega \frac{v_T}{r} $ due to a typo, but the correct equation valid only in radians is $ \omega = \frac{v_T}{r} $).

(Note: If the fourth option is $ \omega = \frac{v_T}{r} $ (after correcting the typo), that’s the answer. Alternatively, the rotational kinematic equation $ \theta = \omega_o t + \frac{1}{2}\alpha t^2 $ is valid for any unit, but the key is $ \omega = \frac{v_T}{r} $.)

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