Equations of Non-Uniform Circular Motion

A body has non-uniform circular motion when its trajectory is a circumference and its angular acceleration is constant. In this section, we are going to study:

• The formulas that correspond to this type of motion
• The relationship between linear quantities and angular quantities
• How to deduce the non-uniform circular motion equations

Non-uniform circular motion equations

Although the former are the main equations of the non uniform circular motion, and the only ones necessary to solve the exercises, it is sometimes useful to know the following expression:

$${\omega }^2={\omega }_0^2+2·\alpha ·∆\phi$$

The previous formula allows us to relate velocity and angular distance traveled if the acceleration is known, and can be deduced from the previous ones, as you can see next.

$$\left\{\begin{array}{l}\omega ={\omega }_0+\alpha ·t\\ \phi ={\phi }_0+{\omega }_0·t+\frac{1}{2}·\alpha ·t^2\end{array}\right.\Rightarrow \left\{\begin{array}{l}t=\frac{\omega -{\omega }_0}{\alpha }\\ ∆\phi ={\omega }_0·t+\frac{1}{2}·\alpha ·t^2\Rightarrow ∆\phi ={\omega }_0\left(\frac{\omega -{\omega }_0}{\alpha }\right)+\frac{1}{2}·\alpha ·{\left(\frac{\omega -{\omega }_0}{\alpha }\right)}^2\end{array}\right.;$$

$$2·\alpha ·∆\phi ={\omega }^2-{\omega }_0^2$$

Relationship between angular and linear quantities

The non-u.c.m. is a circular motion, and as such, the angular and linear quantities are related through the radio R.

From the previous table the following linear quantities can be easily deduced:

$$\begin{gathered} s=\phi ·R={\phi }_0·R+{\omega }_0·R·t+\frac{1}{2}·\alpha ·R·t^2\Rightarrow \\ \Rightarrow \boxed{s=s_0+v_0·t+\frac{1}{2}·a_t·t^2} \end{gathered}$$

$$\begin{gathered} v=\omega ·R={\omega }_0·R+\alpha ·R·t\Rightarrow \\ \Rightarrow \boxed{v=v_0+a_t·t} \end{gathered}$$

$$\begin{gathered} a_t=\alpha ·R=\text{constant}·R\Rightarrow \\ \Rightarrow \boxed{a_t=\text{constant}} \end{gathered}$$

Finally, remember that the total acceleration of a body can be expressed as a function of its intrinsic components, giving us its magnitude:

$$a=\sqrt{{a_t}^2+{a_n}^2}$$

Deduction of equations

To obtain the equations of non-uniform circular motion we proceed in the same way as we did with the uniformly accelerated linear motion (u.a.r.m.), but considering angular magnitudes, rather than linear. We will consider the following properties:

• Angular acceleration is constant ($\alpha =\text{constant}$)
• Additionally, this implies that the average and instantaneous angular velocities of the motion always have the same value

It is, therefore, about determining an expression for the angular velocity, and an another for the angular position (we already know that the angular acceleration is constant). Considering the above restrictions, we get:

$$\alpha ={\alpha }_m=\frac{∆\omega }{∆t}=\frac{\omega -{\omega }_0}{t}\Rightarrow \omega ={\omega }_0+\alpha ·t$$

This first equation relates the angular velocity of the body with its angular acceleration at any moment in time and it is a straight line (ω) whose slope coincides with the angular acceleration and whose y-coordinate at the origin is the initial angular velocity (ω₀). We need to get an equation that allows us to obtain the position. There are different methods to deduce it. We will use the Merton´s theorem: already used in the u.a.r.m., which allows us to say that the angle traveled in a non-uniform circular motion, matches the corresponding to a u.c.m. with angular velocity equal to the arithmetic average of the final and initial angular velocities of the time interval under consideration.

$$\begin{gathered} \left.\begin{array}{r}{\omega }_m=\frac{\omega +{\omega }_0}{2}\\ ∆\phi =\phi -{\phi }_0={\omega }_m·∆t\end{array}\right\}\begin{array}{c}\Rightarrow \end{array}\phi -{\phi }_0=\frac{\omega +{\omega }_0}{2}·∆t\underset{\left[1\right] y \left[2\right]}{\Rightarrow }\begin{array}{c}\Rightarrow \end{array}\phi -{\phi }_0=\frac{{\omega }_0+\alpha ·t+{\omega }_0}{2}·t\Rightarrow \\ \Rightarrow \phi ={\phi }_0+{\omega }_0·t+\frac{1}{2}·\alpha ·t^2 \end{gathered}$$

Where have we applied:

$$\begin{array}{c}\left[1\right] \omega ={\omega }_0+\alpha ·t\\ \left[2\right]∆t=t-t_0\underset{t_0=0}{=}t\end{array}$$