Question

find expression for velocity and acceleration of a moving particle in plane polar coordinates​

Answer 1

Answer:

Explanation:

Velocity And Acceleration In Cylindrical Coordinates

Velocity of a physical object can be obtained by the change in an object's position in respect to time.

Generally, x, y, and z are used in Cartesian coordinates and these are replaced by r, θ, and z.

Answer 2

For a moving particle, the velocity is $\vec{v} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}$ while the acceleration is $\vec{a} = (\ddot{r}-r\dot{\theta^2}) \hat{r} + (r \ddot{\theta}+ 2 \dot{r} \dot{\theta}) \hat{\theta}$ in the plane polar coordinates.

Deriving expression for velocity

In the cartesian coordinate system, $\vec{r} = x \hat{i} + y \hat{j}$ . . . . . . . . (1)

And, for the plane polar coordinate, the position vector $\vec{r} = r \cdot \hat{r}$  

Therefore, by coordinate transformation;

$\vec{r} = r cos(\theta) \hat{i} + r sin(\theta) \hat{j}$

And,

$\hat{r} = \frac{\partial \vec{r}/ \partial r}{|\partial \vec{r}/ \partial r|} = cos(\theta) \hat{i} + sin(\theta) \hat{j} \\\Rightarrow \dot{\hat{r}} = \dot{\theta}(-sin(\theta) \hat{i} + cos(\theta) \hat{j}) = \dot{\theta} \hat{\theta}$   . . . . . . . (2)

$\hat{\theta} = \frac{\partial \vec{r}/ \partial \theta}{|\partial \vec{r}/ \partial \theta|} = - sin(\theta) \hat{i} + cos(\theta) \hat{j}\\\Rightarrow \dot{\hat{\theta}} = -\dot{\theta}(sin(\theta) \hat{i} + cos(\theta) \hat{j}) = -\dot{\theta} \hat{r}$ . . . . . . . (3)

Since, $\vec{v} = \frac{d\vec{r}}{dt}$, therefore, $\vec{v} = \frac{d\vec{r}}{dt} = \frac{(r \cdot \hat{r})}{dt} = \dot{r} \hat{r} + r \dot{\hat{r}}$

from (2), we will get

$\vec{v} = \dot{r} \hat{r} + r \dot{\theta}\hat{\theta}$ . . . . . . . (4)

Deriving expression for acceleration

Since $\vec{a} = \frac{d\vec{v}}{dt}$

therefore, $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d(\dot{r} \hat{r} + r \dot{\theta}\hat{\theta})}{dt} = \frac{d(\dot{r} \hat{r})}{dt} + \frac{d(r \dot{\theta}\hat{\theta})}{dt}$ . . . . . . . (5)

Solving (5) further, you will get $\vec{a} = (\ddot{r}-r\dot{\theta^2}) \hat{r} + (r \ddot{\theta}+ 2 \dot{r} \dot{\theta}) \hat{\theta}$

Hence, for a moving particle, the velocity is $\vec{v} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}$ while the acceleration is $\vec{a} = (\ddot{r}-r\dot{\theta^2}) \hat{r} + (r \ddot{\theta}+ 2 \dot{r} \dot{\theta}) \hat{\theta}$ in the plane polar coordinates where the $\vec{r} = r \cdot \hat{r}$ is the position vector of the particle.