Consider a particle moving along a circular path of constant radius r. If its motion is nonuniform, then its angular speed ω and linear speed v both change with time.
Therefore, in general, the particle has both angular acceleration \(\vec{a}\) = \(\frac{d \vec\omega}{dt}\) and tangential acceleration \(\vec{a_t}\). \(\vec{a}\) has the direction of \(d\vec \omega\), which is in the direction of \(\vec \omega\) if \(\omega\) is increasing and opposite to \(\vec \omega\) if \(\omega\) is decreasing.
At any instant, that linear velocity \(\vec{v}\), angular velocity \(\vec \omega\) and radius vector \(\vec r\) are related by
\(\vec{v}\) = \(\vec \omega\) x \(\vec r\) ...(1)
The linear acceleration of the particle is
\(\vec{a}\) = \(\frac{d \vec v}{dt}\)......(2)
\(\vec{a}\) x \(\vec{r}\) is tangential to the circular path and is in the direction of \(\vec{v}\) if \(\vec{a}\) in the direction of \(\vec \omega\), and it is opposite to \(\vec{v}\) if \(\vec{a}\) is opposite to \(\vec \omega\). Thus, \(\vec{a}\) x \(\vec{r}\)is the tangential acceleration, \(\vec{a_t}\).
\(\vec{a_t}\) = \(\vec{a}\) x \(\vec{r}\)....(4)
In magnitude, \({a_t}\) = ar
Since \(\vec{a}\) is perpendicular to \(\vec{r}\).
Also, \(\vec \omega\) x \(\vec{v}\) is along the radius toward the centre of the circle, i.e, opposite to \(\vec{r}\), i.e, along - \(\vec{r}\); this acceleration is called the radial or centripetal acceleration \(\vec{a_r}\).
\(\vec{a_r}\) = \(\vec \omega\) x \(\vec{v}\)....(5)
In magnitude, ar = ωv
since \(\vec \omega\) is perpendicular to \(\vec{v}\).
∴ \(\vec{a}\)= \(\vec{a_t}\) + \(\vec{a_r}\)....(6)
This is the required expression.
