Relationship between Motion Variables
Besides understanding the definitions and corresponding formulas for linear motion quantities and angular quantities, we should also be aware of the relationship between these quantities. Velocity is the first derivative of displacement with respect to time, \( v=\frac{dx}{dt} \), while acceleration is the first derivative of velocity with respect to time, \( a=\frac{dv}{dt} \) as well as the second derivative of displacement \( a=\frac{d^2{x}}{dt^2}. \) Likewise, angular velocity is the first derivative of angular displacement with respect to time, \( \omega=\frac{d\theta}{dt} \), while angular acceleration is the first derivative of angular velocity with respect to time, \( \alpha=\frac{d\omega}{dt} \) as well as the second derivative of angular displacement \( \alpha=\frac{d^2\theta}{dt^2}. \) Lets check our understanding with the following two examples, starting with angular motion.
Given the position function, \( x(t)=\left(9\;\mathrm{\frac{m}{s^2}}\right) t^2 +\left(6\;\mathrm{\frac{m}{s^2}}\right)t -4\;\mathrm{m}, \) calculate the velocity and acceleration functions.
Solution
To find the velocity function we can calculate the derivative of the position function with respect to time.
\begin{align}v(t)&=\frac{\mathrm{d}x}{\mathrm{d}t}\\v(t)&= \frac{\mathrm{d}}{\mathrm{d}t} \left( \left(9;\mathrm{\frac{m}{s^2}}\right) t^2 +\left(6\;\mathrm{\frac{m}{s}}\right)t -4\;\mathrm{m}\right)\\v(t)&=\left(18\;\mathrm{\frac{m}{s^2}}\right)t+6\;\mathrm{\frac{m}{s}}\end{align}
Similarly, since acceleration is the second derivative of the position function, we can find it by calculating the derivative of the velocity function.
\begin{align}\\a(t)&=\frac{\mathrm{d}v}{\mathrm{d}t}\\a(t)&= \frac{\mathrm{d}}{\mathrm{d}t} \left(\left(18\;\mathrm{\frac{m}{s^2}}\right)t+6\;\mathrm{\frac{m}{s}}\right)\\a(t)&=18\;\mathrm{\frac{m}{s^2}}\\\end{align}
One more time for angular motion.
Given the angular position function, \( \theta(t)=\left(5\;\mathrm{\frac{rad}{s^2}}\right) t^2 +\left(8\;\mathrm{\frac{rad}{s^2}}\right)t -6\;\mathrm{rad}, \) calculate the angular velocity and angular acceleration functions.
Solution
To find the angular velocity function we can calculate the derivative of the angular position function with respect to time.
\begin{align}\omega(t)&=\frac{\mathrm{d}\theta}{\mathrm{d}t}\\\omega(t)&= \frac{\mathrm{d}}{\mathrm{d}t} \left( \left(5\;\mathrm{\frac{rad}{s^2}}\right) t^2 +\left(8\;\mathrm{\frac{rad}{s}}\right)t -6\;\mathrm{rad}\right)\\\omega(t)&=\left(10\;\mathrm{\frac{rad}{s^2}}\right)t+8\;\mathrm{\frac{rad}{s}}\end{align}
Similarly, since angular acceleration is the second derivative of the angular position function, we can find it by calculating the derivative of the angular velocity function.
\begin{align}\\\alpha(t)&=\frac{\mathrm{d}\omega}{\mathrm{d}t}\\\alpha(t)&= \frac{\mathrm{d}}{\mathrm{d}t} \left(\left(10\;\mathrm{\frac{rad}{s^2}}\right)t+8\;\mathrm{\frac{rad}{s}}\right)\\\alpha(t)&=10\;\mathrm{\frac{rad}{s^2}}\\\end{align}
Now that we understand the relationship between the quantities themselves, we also need to discuss how linear motion quantities and angular motion quantities are related to one another. These relationships appear in the table below, where \( r \) is the radius of the circular motion.
Variable | Linear Abbreviation | Linear SI units | Angular Abbreviation | Angular SI units | Relationship |
acceleration | $$a$$ | $$\frac{m}{s^2}$$ | $$\alpha$$ | $$\frac{rad}{s^2}$$ | $$a=\alpha{r}, \alpha=\frac{a}{r}$$ |
velocity | $$v$$ | $$\frac{m}{s}$$ | $$\omega$$ | $$\frac{rad}{s}$$ | $$v=\omega{r}, \omega=\frac{v}{r}$$ |
displacement | $$x$$ | $$m$$ | $$\Delta \theta$$ | $$rad$$ | $$x=\theta{r}, \theta=\frac{x}{r}$$ |
time | $$t$$ | $$s$$ | $$t$$ | $$s$$ | $$t=t$$ |
To better understand these relationships, let us look at the diagram below.
Fig. 1 - A diagram illustrating the relationship between linear kinematic variables and angular kinematic variables where \( v=\text{velocity},\) \(r=\text{radius},\) \( X=\text{linear displacement},\) \( \Delta \theta=\text{angular displacement},\) and \( \omega=\text{angular velocity.} \)
Angular displacement, \(\theta\) is a measurement of how much a body has revolved around a specified axis, given as an angle usually in units of radians. Angular displacement is directly related to linear displacement, as linear displacement concerns the straight-line distance between points while angular displacement concerns the curved motion path between these points. Linear displacement is proportional to the radius of rotation and angular displacement. Tangential velocity, \(v\) describes the instantaneous linear component of an object's motion relative to its circular path in the direction perpendicular to the normal to the direction to the centre of rotation. A way to visualize this is to imagine spinning a mass on a string - if we cut the string, the mass will continue moving with its tangential velocity at the instant the string was severed. This direction is perpendicular to the centre of the circle of rotation of the mass.
Similarly, this relationship describes the relationship between angular acceleration and linear acceleration. If we insert the formula for angular velocity, \( \omega=\frac{v}{r} \), into the equation for angular acceleration, \( \alpha=\frac{\Delta{\omega}}{\Delta{t}} \), we can derive the corresponding equation that relates angular acceleration to the instantaneous linear acceleration, \( \alpha=ar. \) Linear motion quantities are analogous to angular motion quantities.
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