Physics, asked by narottam272, 1 month ago

Obtain expressions for the velocity and acceleration of a particle in terms of plane polar coordinate systems.​

Answers

Answered by meenababli12
0

Answer:

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Answered by bharathparasad577
0

Answer:

Explanation:

Concept:

The system of polar coordinates. The polar axis is a line segment that extends to the right from the graph's center and is known as the positive x-axis in the Cartesian coordinate system. The pole, or origin, of the coordinate system, which is the center point, is equal to r=0.
Solution:

Velocity and Acceleration in Polar Coordinate:

Velocity Vector in Polar Coordinate:

  $\vec{v}=\frac{d(r \hat{r})}{d t}=\frac{d r}{d t} . \hat{r}+r \frac{d \hat{r}}{d t}=\dot{r} \hat{\mathbf{r}}+r \frac{d \hat{\mathbf{r}}}{d t} \Rightarrow \vec{v}=\dot{r} \hat{r}+r \dot{\theta} \hat{\theta}$$

where $\dot{r}$ is radial velocity in $\hat{r}$ direction and $r \dot{\theta}$ is tangential velocity in $\hat{\theta}$ direction as shown in figure and the magnitude to velocity vector $|v|=\sqrt{\dot{r}^{2}+r^{2} \dot{\theta}^{2}}$

Acceleration Vector in Polar Coordinate:

 $$\begin{aligned}&\frac{d \vec{v}}{d t}=\frac{d \dot{r}}{d t} \hat{r}+\dot{r} \frac{d \hat{r}}{d t}+\frac{d r}{d t} \dot{\theta} \hat{\theta}+r \frac{d \dot{\theta}}{d t} \hat{\theta}+r \dot{\theta} \frac{d \hat{\theta}}{d t} \\\\&\frac{d \vec{v}}{d t}=\ddot{r} \hat{r}+\dot{r} \dot{\theta} \hat{\theta}+\dot{r} \dot{\theta} \hat{\theta}+r \ddot{\theta} \hat{\theta}+r \dot{\theta}(-\dot{\theta}) \hat{r} \\\end{aligned}$$

&\vec{a}=\left(\ddot{r}-r \dot{\theta}^{2}\right) \hat{r}+(r \ddot{\theta}+2 \dot{r} \dot{\theta}) \hat{\theta} \Rightarrow \vec{a}=a_{r} \hat{r}+a_{\theta} \hat{\theta}

$a_{r}=\ddot{r}-r \dot{\theta}^{2}$ is radial acceleration and $a_{\theta}=r \ddot{\theta}+2 \dot{r} \dot{\theta}$ is tangential acceleration .

So Newton's law in polar coordinate can be written as

F_{r}=m a_{r}=m\left(\ddot{r}-r \dot{\theta}^{2}\right)$ where $F_{r}$ is external force in radial direction.

$F_{\theta}=m a_{\theta}=m(r \ddot{\theta}+2 \dot{r} \dot{\theta})$

where $F_{\theta}$ is external force in tangential direction.

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