Connecting Linear and Rotational Motion

In physics, there is a goal to unify all theories. This has been accomplished for predicting the motion of macroscopic objects. As we have discussed previously, different theories and laws have been combined to describe how forces affect the position, velocity, and acceleration of an object that is either rotating or moving translationally. In this article, we will discuss how to connect translational and rotational motion, as there are objects that rotate and move linearly at the same time.

Fig. 1 - A spinning top undergoing linear and rotational motion.

Angular Displacement, Velocity, and Acceleration

In a previous article, we defined the equations to determine an object's instantaneous velocity and acceleration using calculus. These definitions have a rotational analogous. Angular displacement is the angle by which a point in a rigid system rotates around an axis. Rigid systems are systems that hold their shape, but different points within the system can move in different directions. If the rotation of the system is about the center of mass, the rigid system can be treated as an object. For example, the rotation of the Earth about its axis is negligible if we treat the Earth as a point, a center of mass rotating around the Sun. The angular velocity is the rate of change of the angular displacement, while the angular acceleration is the rate of change of the angular velocity. We will present the calculus definitions below.

$$\begin{array}{rcl}\Delta x=x_\text{f}-x_\text{i}&\leftrightarrow&\Delta \theta=\theta_\text{f}-\theta_\text{i}\\v=\frac{\operatorname dx}{\operatorname dt}&\leftrightarrow&\omega=\frac{\operatorname d\theta}{\operatorname dt}\\a=\frac{\operatorname dv}{\operatorname dt}&\leftrightarrow&\alpha=\frac{\operatorname d\omega}{\operatorname dt}\end{array}\\$$

You can also use intervals to determine the average angular velocity and acceleration.

$$\begin{align*}\alpha&=\frac{\mathrm d\omega}{\mathrm dt},\\\displaystyle\int_{{\mathrm\omega}_0}^{\mathrm\omega}\mathrm d\omega&=\int_0^t\alpha\;\mathrm dt\;,\\\omega&=\omega_0+\alpha t,\end{align*}$$

where \(\omega_0\) is the constant of integration that represents the initial angular velocity in radians per second \((\frac{\mathrm{rad}}{\mathrm s})\). The equation above is useful for physics problems that do not provide or ask you to find the angular displacement. If we integrate once again, we can find the equation for the angular position and displacement of a rotating object:

$$\begin{align*}\omega&=\frac{\mathrm{dθ}}{\mathrm dt},\\\omega&=\omega_0+\alpha t,\\\frac{\mathrm{dθ}}{\mathrm dt}&=\omega_0+\alpha t,\\\int_{\theta_0}^\theta\mathrm{dθ}&=\int_0^t\left(\omega_0+\alpha t\right)\mathrm dt,\\\theta&=\theta_0+\omega_0t+\frac12\alpha t^2.\end{align*}\\$$

The equation above is useful for physics problems that do not provide or ask you to find the angular velocity. We can derive one last useful equation by completing the square. In case you don't remember how to complete the square, we will show you what you need:

$$ax^2+bx+c\leftrightarrow a{(x+d)}^2+e,$$

where \(d=\frac b{2a}\) and \(e=c-\frac{b^2}{4a}\). In our case, \(d=\frac{\omega_0}\alpha\) and \(e=\theta_0-\frac{\omega_0^2}{2\alpha}\). Now we can rewrite the equation for the angular position of the object,

$$\theta=\theta_0+\frac12\alpha{(t+\frac{\omega_0}\alpha)}^2\;-\frac{\omega_0^2}{2\alpha}.$$

If we rearrange the equation for the angular velocity and substitute the term \(t+\frac{\omega_0}\alpha\) in the above expression, we obtain a useful expression to solve physics problems that do not provide or ask you to find the time interval:

$$\begin{align*}\omega&=\omega_0+\alpha t,\\\frac\omega\alpha&=t+\frac{\omega_0}\alpha.\end{align*}$$

We substitute this expression in the equation for the angular displacement,

$$\begin{align*}\theta&=\theta_0+\frac12\alpha\left(\frac\omega\alpha\right)^2-\frac{\omega_0^2}{2\alpha},\\\theta&=\theta_0+\frac{\omega^2-\omega_0^2}{2\alpha},\\\omega^2&=\omega_0^2+2\alpha\left(\theta-\theta_0\right).\end{align*}$$

We have proved that equations for uniform acceleration motion in one dimension are identical for translational and rotational motion. The angular displacement, velocity, and acceleration are the rotational analogous to the translational displacement, velocity, and acceleration.

Now we can describe the translational motion of a point rotating around an axis and the rotational motion of a point that is also moving translationally. In the image below we can see how to identify a system that is undergoing translational motion, rotational motion or both at the same time. It is important to note that translational motion can be linear or nonlinear.

Fig. 2 - Visualizing the combination of rotational and translational motion. We also see that translational motion can be linear or nonlinear.

Connecting Translational and Rotational Motion

To connect the angular displacement to the translation of a point in a rigid system, we must first imagine a disk with a given radius that we will call reference line. In a rigid body, all the points along a radial line will have the same angular displacement.

Fig. 3 - Visualization of the reference line to connect translational displacement with angular displacement. In a rigid system, all points along a radial line always have the same angular displacement.

From the image above, we see that if multiply the angular displacement by the length of the radial line, we can determine the arc length. The arc length definition is key to connecting translational and rotational motion,

$$\begin{align*}\Delta s\;&=r\Delta\theta,\\\frac{\Delta s}r&=\Delta\theta.\;\end{align*}$$

Similarly, we can rearrange the definition of the arc length to find the expression for the angular velocity. We must divide both sides of the equation by a time interval,

$$\begin{align*}\Delta s&=r\Delta\theta,\\\frac{\displaystyle\Delta s}{\Delta t}&=r\frac{\displaystyle\Delta\theta}{\Delta t},\;\\\overline v&=r\overline\omega.\end{align*}$$

To find the instantaneous translational and angular velocity, we must use their calculus definitions:

$$\begin{align*}\frac{\displaystyle\operatorname ds}{\operatorname dt}&=r\frac{\displaystyle\operatorname d\theta}{\operatorname dt},\\v&=r\omega.\;\end{align*}$$

We have connected the translational displacement and velocity with the angular displacement and velocity. Also, we can notice that the angular velocity is proportional to the length of the radial line. This means that points that are far away from the rotational axis will move faster than points which are closer to the axis.

To connect the translational acceleration to the angular acceleration, we must differentiate the above expression with respect to time:

$$\begin{align*}\frac{\displaystyle\operatorname dv}{\operatorname dt}&=r\frac{\displaystyle\operatorname d\omega}{\operatorname dt},\\a&=r\alpha.\end{align*}$$

We have just proved that acceleration can emerge from a change in speed caused by angular acceleration.

It is important to mention that this acceleration is different from the centripetal acceleration caused by translational and circular motion. The centripetal acceleration does not produce a change in speed, but a change in velocity because the direction is constantly changing.