P01-240 – Instantaneous Angular Acceleration

Instantaneous Angular Acceleration

In the same way we derived the instantaneous angular velocity from the average angular velocity, we now derive the instantaneous angular acceleration from the average angular acceleration by letting $\mathrm{\Delta }t\to 0$.

With the help from calculus, we define the instantaneous angular acceleration as the following limit

\boxed{\alpha ={\mathop{\mathrm{lim}}_{\mathrm{\Delta }t\to 0} \frac{\mathrm{\Delta }\omega }{\mathrm{\Delta }t}\ }=\frac{d\omega }{dt}\ \ \ \ \ \ \ \ \left(rad/s^2\right)}

where such a limit is, by definition, equal to the derivative of $\omega $ with respect to $t$ and is denoted by $\displaystyle{\frac{d\omega }{dt}}$.

Note: it is common to simply call the instantaneous angular acceleration of a particle its angular acceleration.

Example: consider a particle moving along a circular path with an angular velocity given by $\omega \left(t\right)=-2t+3$. Its instantaneous angular velocity is then given by

\alpha \left(t\right)=-2\ \ \ \ \ \left(rad/s^2\right)

This yields a function of time for $\omega \left(t\right)$ which will output the instantaneous angular velocity of the particle at any instant $t$ of its motion. In particular, a constant value of $-2\ \ rad/s^2$ indicates that every second, the angular speed of the particle decreases by $2\ rad/s$.