I have seen the derivation of relation between velocity and angular velocity in circular motion as $$ S=\theta R, $$ where $S$ is the arc length and $R$ is the radius.

Take a derivative with respect to time on both sides and you get \begin{align} \frac{\mathrm dS}{\mathrm dt} &= R \frac{\mathrm d\theta}{\mathrm dt} , \quad \text{where}\\ \frac{\mathrm dS}{\mathrm dt} &= \text{velocity} \\ \frac{\mathrm d\theta}{\mathrm dt} &= \text{angular velocity}, \quad \text{so}\\ v &= r\omega. \end{align}

My question is $S=\theta R$, in this $S$ is distance (arc) and not displacement, so how is $\mathrm dS/\mathrm dt$ the velocity? It should be speed. Can somebody help me?

  • 1
    $\begingroup$ The v in your equation is the speed not the velocity and it is also angular speed in the equation . The velocity is given by this equation $\vec v = \vec \omega \times \vec r$ $\endgroup$ – Farcher Jun 11 '18 at 16:45
  • $\begingroup$ Dear Mr Farcher , In the above equation dtheta/ dt has been rightly defined as angular velocity as i understand and its not angular speed.Please guide $\endgroup$ – pik selvan Jun 11 '18 at 17:07
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    $\begingroup$ I see that the "velocity vs. speed" issue keeps causing problems. As it's been told here for long, this only happens in English language. When you talk about "force", you can mean both $\vec{F}$ or $|\vec{F}|$, and there's no problem. Why should velocity be different? It's jsut a caprice. You only have to think if you are referring to a vector or referring to a scalar. $\endgroup$ – FGSUZ Jun 11 '18 at 17:56
  • $\begingroup$ $d\theta/dt=\omega$ is the angular "speed" if you want, and $\vec{\omega}$ is a vector whose modulus is $\omega$ and it is perpendicular to the plane of movement. $\endgroup$ – FGSUZ Jun 11 '18 at 17:57
  • $\begingroup$ You are considering circular motion about a fixed axis. Because the direction of the axis is fixed in space, the concepts "angular velocity is a vector fixed in space, perpendicular to the circular motion" and "angular velocity is the rate of change of your angle $\theta$" amount to the same thing. In the more general situation where the axis of rotation is changing direction, this is not so simple - a physical example would be a spinning top that is "wobbling" as it slows down and topples over. $\endgroup$ – alephzero Jun 11 '18 at 21:56

so how is dS/dt the velocity? It should be speed.

For, e.g., a particle in uniform circular motion a distance $R$ from the origin, the position vector $\mathbf{s}$ is given by

$$\mathbf{s} = R\cos(\omega t + \theta_0)\hat{\mathbf{x}} + R\sin(\omega t + \theta_0)\hat{\mathbf{y}}$$

The velocity vector $\mathbf{v}$ is then given by

$$\mathbf{v} = \frac{d\mathbf{s}}{dt} = -R\omega\sin(\omega t + \theta_0)\hat{\mathbf{x}} + R\omega\cos(\omega t + \theta_0)\hat{\mathbf{y}}$$

with magnitude

$$|\mathbf{v}| = \omega R$$

The distance $ds$ the particle travels in time $dt$ is given by

$$ds = R\omega\,dt$$ and so the speed of the particle is, as you note,

$$\frac{ds}{dt} = \omega R$$


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