# Non Uniform Circular Motion

## 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 $$\left( {{F}_{Net}} \right)=ma=m\sqrt{a_{c}^{2}+a_{t}^{2}}$$