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I understand that distance is always a postive number, so its derivative, speed, is also always positive. However, for rotational motion with angles, it seems that $\Delta \theta$ can have a sign, clockwise negative and counterclockwise positive.

  1. When we differentiate $s = \theta r$ ($s$ being the arc length, so therefore it's a distance not a displacement), $\frac{ds}{dt}$ must be speed (distance over time), but $\frac{d\theta}{dt}$ will give angular velocity, which can be negative since $\Delta \theta$ can be negative. However, according to my knowledge, speed cannot be negative, so $v = \omega r$ does not make sense to me.
  2. Even worse, if I blindly ignore this and follow the textbook and differentiate this again with respect to time, it says I get $a = \alpha r$, which is even more confusing because i just differentiated speed over time, which I know can't be an acceleration.

What am I missing here?

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    $\begingroup$ You are missing the difference between speed and velocity. $\endgroup$
    – mike stone
    Apr 18 at 13:26
  • $\begingroup$ By my knowledge, velocity is a displacement vector differentiated over time, and speed is its magnitude. I still get the same problem. $\endgroup$
    – mikeeei
    Apr 18 at 13:27
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    $\begingroup$ Velocity is ${\bf v}=d{\bf x}/dt$ and even in one dimension $dx$ can be negative just as $d\theta$ can be negative. In $s=\theta r$ both $s$ and $\theta$ can have either sign. This why $\omega$ is called the angular velocity not the angular speed. $\endgroup$
    – mike stone
    Apr 18 at 13:30
  • $\begingroup$ I understand, but this does not seem to answer my original question. So is ds/dt in fact a velocity and not speed (which I originally assumed it to be?) $\endgroup$
    – mikeeei
    Apr 18 at 13:32
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    $\begingroup$ Are you comparing motion along the arc of a circle and motion on a straight line chord between two points on a circle? If so you are really trying to describe what an apple tastes like by comparing it to the taste of an orange. Constrain your circular motion to a circle. Note that motion in a circle can be described using the rectilinear concepts, but the math is a lot more complicated than what you are trying to do. $\endgroup$
    – garyp
    Apr 18 at 14:05
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Since s is a vector ds/dt or velocty v is a vector, so it has direction in two or three space, the same with angular velocity . When you write 𝑎=𝛼𝑟 a, 𝛼 and r are usually vectors, it seems you look at very simple cases only, where all the vectors have only one direction.

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  • $\begingroup$ 𝑠 = 𝜃𝑟, so how can arc length be a vector? $\endgroup$
    – mikeeei
    Apr 18 at 13:52
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It's easiest to work in polar coordinates with the basis unit vectors $\hat{r}$ and $\hat{\theta}$. Then there is no confusion about positive and negative signs-- the particle's position, velocity, and acceleration are firmly defined as vectors. For general motion in a plane (not necessarily circular, $r$ can vary with time), you can show that $$\vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$$ and $$\vec{a} = (-r\dot{\theta}^2+\ddot{r})\hat{r} + (r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}$$ For a detailed derivation check out this link (pg. 32): https://www.ucl.ac.uk/~zcapd49/phas1247coursenotes.pdf

When $r$ is held constant, we get equations of the same form as those you mentioned: $$\vec{v}=r\dot{\theta}\hat{\theta}$$ and $$\vec{a}=-r\dot{\theta}^2\hat{r} + r\ddot{\theta}\hat{\theta}.$$ So we see that $r\alpha$ is only the tangential component of the acceleration.

Notes:

  • Just in case you aren't familiar with it, a dot over a variable is another notation for the time derivative of the variable, e.g., $\dot{r} = \frac{\,dr}{\,dt}$. Similarly for a double dot: $\ddot{r}=\frac{\,d^2r}{\,dt^2}$.
  • $\hat{r}$ is the radial unit vector; it points outward from the origin to the position of the particle and has magnitude one. $\hat{\theta}$ is the counterclockwise tangential unit vector. It points perpendicular to the line connecting the particle's position with the origin, in a ccw sense. As with all unit vectors, it has magnitude one. (If you already know this, just ignore it.)
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See this answer for a detailed discussion of the following kinematical analysis in the context of orbital motion.

$\underline{\textit{Analysis:}}$

Let $\vec{r}$ be the displacement of the material particle $P$ from the material particle designated to be the origin, $\vec{v}^R_P:=\vec{v} := \frac{{}^Rd}{dt}\vec{r} := \dot{\vec{r}}$ and $\vec{a}_P^R:=\vec{a} := \frac{{}^Rd^2}{dt^2}\vec{r} := \ddot{\vec{r}}$ be the velocity and acceleration of the material particle calculated using the reference frame $R$, respectively. Further, we denote the time derivative of a vector calculated using the reference frame $R$ by $\dot{(\cdot)}:=\frac{{}^Rd}{dt}(\cdot)$. Denoting $r:=|\vec{r}|$ and $\hat{r} := \frac{\vec{r}}{|\vec{r}|}$, we can see that $\dot{|\hat{r}|} := \frac{d |\hat{r}|}{dt} = \dot{1} = 0 = \frac{1}{2} \hat{r} \cdot \dot{\hat{r}}$, which implies that $\dot{\hat{r}} \perp \hat{r}$. Further, $\vec{v}=\dot{r}\hat{r}+r\dot{\hat{r}}$ and $\vec{a}=\ddot{r}\hat{r}+2\dot{r}\dot{\hat{r}}+r\ddot{\hat{r}}$.

Let us calculate the time derivative using a reference frame $M$ rotating but not translating w.r.t. the reference frame $R$ such that an axis of $M$ always passes through the material particle and let us denote the angular velocity and speed of $M$ w.r.t. $R$ by $\vec{\omega}^{MR}:=\vec{\omega}$ and $\frac{{}^Rd}{dt}\vec{\omega}^{MR}:=\vec{\alpha}$ respectively. Then, the Coriolis' theorem implies that $\vec{v}_P^R:=\vec{v}=\frac{{}^Md}{dt}\vec{r}+\vec{\omega}\times\vec{r}=\dot{r}\hat{r}+\vec{\omega}\times r\hat{r}$. Since $\vec{v}=\dot{r}\hat{r}+r\dot{\hat{r}}$, we obtain that $\vec{\omega}\times \hat{r}=\dot{\hat{r}}$ so that the unit vectors $\hat{r},\frac{\dot{\hat{r}}}{\text{norm}(\dot{\hat{r}})},\frac{\vec{\omega}}{|\vec{\omega}|}$ constitute a unit right handed traid (which can be) associated with the reference frame $M$ (this reference frame is instantaneously defined at each time instant in the duration of motion of $P$ and mimics a rigid body although no such real-world rigid body may be associated with it).

Let us now assume that the motion of $P$ is planar. Due to the assumption of the planar motion, the vector $\frac{\vec{\omega}}{|\vec{\omega}|}$ is identical to the constant unit vector $\hat{k}$ which is normal to the plane of motion determined by the linearly independent pair $\hat{r},\dot{\hat{r}}$ and we can write $\vec{\omega}=\dot{\theta}\hat{k}$ where $\theta$ denotes the angular position of $P$ (which is identical to the scalar angular displacement of the reference frame $M$ w.r.t. the reference frame $R$) in it's plane of motion. Therefore, $\vec{v}=\dot{r}\hat{r}+r\dot{\hat{r}}=\dot{r}\hat{r}+r\dot{\theta}\frac{\dot{\hat{r}}}{\text{norm}(\dot{\hat{r}})}$ so that $\text{norm}(\dot{\hat{r}})=\dot{\theta}$ and $\vec{a}=\ddot{r}\hat{r}+2\dot{r}\dot{\hat{r}}+r\ddot{\hat{r}}=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(r\ddot{\theta}+2\dot{\theta}\dot{r})\frac{\dot{\hat{r}}}{\text{norm}(\dot{\hat{r}})}$ so that $\ddot{\hat{r}}=-\dot{\theta}^2\hat{r}+r\ddot{\theta}\frac{\dot{\hat{r}}}{\text{norm}(\dot{\hat{r}})}$. Notice that the unit vector $\frac{\dot{\hat{r}}}{\text{norm}(\dot{\hat{r}})}$ indicating the tangent to the trajectory of P may be denoted by, say $\hat{\theta}$, for brevity.

The case of uniform circular planar motion is obtained by setting $\dot{r}=0$ and $\ddot{\theta}=0$ which corresponds to the case in the OP, wherein the arc length travelled by $P$ is given by $s=r\theta$.


$\underline{\textit{Answers to the questions in the OP:}}$

  1. The variable $v$ referred to in the OP is not the speed but the signed magnitude of the velocity. In other words, consider the force due to gravity on a material particle of mass $m$, expressed in the preferred coordinate system of a reference frame which is oriented such that the constant unit vector $\hat{k}$ is directed vertically downwards, so that the force due to gravity is $\vec{F}_\text{gravity}=-mg\hat{k}$. In this expression, $-mg$ is the signed magnitude or component of the force, while $mg=|\vec{F}_\text{gravity}|:=\|\vec{F}_\text{gravity}\|$ is the magnitude or norm so that we can write $\vec{F}_\text{gravity}=-|\vec{F}_\text{gravity}|\hat{k}$. If we denote ${F}_\text{gravity}:=-mg$, then $\vec{F}_\text{gravity}=-|\vec{F}_\text{gravity}|\hat{k}={F}_\text{gravity}\hat{k}=-mg\hat{k}$. Similarly, if the position vector of the material particle is given by $\vec{r}=r(\cos\theta\hat{i}+\sin\theta\hat{j})$ executing the planar circular motion $\dot{r}\equiv0$ with the angular speed $\omega:=\dot{\theta}$, then the variable $v:=\omega r$ in the OP expresses the velocity of the material particle as $\vec{v}=v(-sin\theta\hat{i}+\cos\theta\hat{j})$. In this case, the arc length or distance travelled by the material particle as a function of the angle $\theta$ is given by $s=r\theta$. Note that $\|\vec{v}\|:=|v|=|\omega r|$.
  2. A similar explanation provides an understanding of the reason why $a=\alpha r$ where $\alpha:=\ddot{\theta}$, since $\vec{a}=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(r\ddot{\theta}+2\dot{\theta}\dot{r})\frac{\dot{\hat{r}}}{\text{norm}(\dot{\hat{r}})}$.

The questions in the OP arise from the conflating the definition of the magnitude or norm of a Eulcidean vector with that of the signed magnitude or components of the vector.


$\underline{\textit{Learning value:}}$

In general, the time derivative of a vector of constant magnitude is perpendicular to the vector. Therefore, $\vec{r} \cdot \dot{\vec{r}} = r \hat{r} \cdot (\dot{r} \hat{r} + r \dot{\hat{r}})$, so that $\vec{r} \cdot \vec{v} = r \dot{r}$. This idea is crucial to developing kinematic relationships in certain analyses.

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