# Correct way to calculate angular displacement from angular velocity

I have a set of data of angular velocity measured at discrete time stamps, and i want to calculate the angular displacement of the rotating object.

Taking into account that $\omega = d\theta/ dt$ ,

is it correct to assume that $\theta_f = \omega*(t_f - t_i) + \theta_i$?

For non constant $\omega$, is it ok to use in the above equation the $\omega = ( \omega_f + \omega_i ) / 2$?

F stands for final time and i stands for initial time.

Does the elementary $\theta = \omega*t$ hold only when $\omega$ is constant?

$$\theta_f = \omega(t_f - t_i) + \theta_i$$

is correct if $\omega$ is constant for the given time interval.

Let's start from basics,

The rate of change of angular displacement ($\theta$) is angular velocity ($\omega$).

$$\frac{\theta_f-\theta_i}{t_f-t_i}=<\omega>$$

This formula gives average angular velocity

$$\lim_{ t_i\rightarrow t_f}\frac{\theta_f-\theta_i}{t_f-t_i}=<\omega>$$

Since,the time interval is so small,instead of calling average velocity over very short time period,we call it instantaneous angular velocity.

$$\frac{\text{d}\theta}{\text{d}t}=\omega$$ This formula gives instantaneous angular velocity

Let's look at an example where $\omega$ is a function of time like

$$\omega=5 t^2$$

$$\frac{\text{d}\theta}{\text{d}t}=\omega=5t^2$$

Let at $t=0$, $\theta=\theta_i$ and $t=t_f$, $\theta=\theta_f$

So,$$\omega_{t=0}=5(0)^2=0$$

$$\omega_{t=f}=5(t_{f})^2$$

If you perform integration to find the net angular displacement,you will see that your method does not equate to the integration result.

• Hello, thank you for the in depth answer. What you state is true, but since my sample ratio is close to 1000 Hz, taking the angular veocity to be constant between two consecutive points is not that harsh of a simplification. Oct 17, 2016 at 16:41