Non Uniform Circular Motion
If the speed of the particle in a horizontal circular motion changes with respect to time, then its motion is said to be non-uniform circular motion. Consider a particle describing a circular path of radius r with center at O. let at an instant the particle be at P and \(\overrightarrow{v}\) be its linear velocity and \(\overrightarrow{\omega }\) be its angular velocity.
Then, \(\overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}……(1)\)
After differentiating equation (1) with respect to time (t) we get:
\(\frac{d\overrightarrow{v}}{dt}=\frac{d\overrightarrow{\omega }}{dt}\times \overrightarrow{r}+\overrightarrow{\omega }\times \frac{d\overrightarrow{r}}{dt}……..(2)\) ; Here,\(\frac{d\overrightarrow{v}}{dt}=\overrightarrow{a}\) (Resultant Acceleration)
\(\overrightarrow{a}=\overrightarrow{\alpha }\times \overrightarrow{r}+\overrightarrow{\omega }\times \overrightarrow{v}\) \(\Rightarrow \frac{d\overrightarrow{\omega }}{dt}=\overrightarrow{\alpha }\) (Angular Acceleration)
\(\overrightarrow{a}=\overrightarrow{{{a}_{t}}}+\overrightarrow{{{a}_{c}}}…….(3)\) ; \(\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}\) (Linear Velocity)
Thus the resultant acceleration of the particle at P has two component accelerations.
1) Tangential Acceleration: \(\left( \overrightarrow{{{a}_{t}}} \right)=\overrightarrow{\alpha }\times \overrightarrow{r}\)
It acts along the tangent to the circular path at P in the plane of circular path. According to right rule since \(\overrightarrow{\alpha }\) and \(\overrightarrow{r}\) are perpendicular to each other, therefore, the magnitude of tangential acceleration is given by:
\(|\overrightarrow{{{a}_{t}}}|=|\overrightarrow{\alpha }\times \overrightarrow{r}|=\alpha r\sin {{90}^{0}}=\alpha r\)2) Centripetal Acceleration: \(\left( \overrightarrow{{{a}_{c}}} \right)=\overrightarrow{\omega }\times \overrightarrow{v}\)
It is also called centripetal acceleration of the particle at P. it acts along the radius of the particle P. according to right hand rule since \(\overrightarrow{\omega }\) and \(\overrightarrow{v}\) are perpendicular to each other, therefore, the magnitude of centripetal acceleration is given by:
\(|\overrightarrow{a}|=|\overrightarrow{\omega}\times \overrightarrow{v}|=\omega v\sin {{90}^{0}}=\omega v=\omega\left(r\omega \right)={{\omega }^{2}}r=\frac{{{v}^{2}}}{r}\)
3) Force: In non-uniform circular motion the particle simultaneously possesses two forces.
Centripetal Force\(\left( {{F}_{c}} \right)=m{{a}_{c}}=\frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}\)
Tangential Force\(\left( {{F}_{t}} \right)=m{{a}_{t}}\)
Net Force \(\left( {{F}_{Net}} \right)=ma=m\sqrt{a_{c}^{2}+a_{t}^{2}}\)