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Is it possible for a particle to have angular velocity but no angular acceleration? Even if the angular velocity does not change, does there always need to be a centripetal / centrifugal acceleration acting on the object?

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  • $\begingroup$ If velocity is constant then acceleration is 0 since accelereation is changes in velocity with respect to time. If there's no change then the slope (acceleration) is 0 $\endgroup$
    – Billy Istiak
    Commented Jan 3, 2022 at 7:22

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To respond to the question in the title:

To make a turn in space requires that the particle's velocity vector change direction. This means the particle is accelerating towards the center of the turn and will hence experience a force acting at right angles to its direction of motion as it turns. That force must be maintained for as long as the particle is in circular motion. The larger the radius of the turn, the smaller will be the acceleration and the force experienced by the particle.

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There exists a one-to-opne correspondent between linear motion and rotational motion. We obtain the formulas for the rotational motion if we replace \begin{align} \textrm{position } r &\to \textrm{angle } \varphi \\ \textrm{velocity } v &\to \textrm{anglular velocity } \omega \\ \textrm{acceleration } a &\to \textrm{anglular acceleration } \alpha \\ \textrm{mass } m &\to \textrm{momentum of inertia } \theta \\ \textrm{(linear) momentum } p &\to \textrm{angular momentum } L \end{align} This correspondent is also true for the time derivatives. E.g. just as $v = \frac{dr}{dt}$ we have $\omega = \frac{d\varphi}{dt}$.

Q1: Is it possible for a particle to have angular velocity but no angular acceleration? Yes, because we have $\alpha = d\omega/dt$. Hence, if $\omega$ is constant we have $\omega \ne 0$ and $\alpha = 0$.

Q2: Even if the angular velocity does not change, does there always need to be a centripetal / centrifugal acceleration acting on the object? If the angular velocity $\omega$ is constant there is no (net) "radial" acceleration $\alpha$ acting on the object. With "radial" I try to emphasise that the angular acceleration is the component which is perpendicular to the linear velocity $v$ of the object.

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