# How exactly are linear and rotational velocity and acceleration related?

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?

• You are missing the difference between speed and velocity. Apr 18, 2021 at 13:26
• By my knowledge, velocity is a displacement vector differentiated over time, and speed is its magnitude. I still get the same problem. Apr 18, 2021 at 13:27
• 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. Apr 18, 2021 at 13:30
• 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?) Apr 18, 2021 at 13:32
• 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. Apr 18, 2021 at 14:05

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.

• 𝑠 = 𝜃𝑟, so how can arc length be a vector? Apr 18, 2021 at 13:52

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.)

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.