Question

The radial component of acceleration in plane polar coordinates is given by:

• A. \( \frac{d^2r}{dt^2} \)
• B. \( \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 \)
• C. \( \frac{d^2r}{dt^2} + r\left(\frac{d\theta}{dt}\right)^2 \)
• D. \( 2r\frac{d\theta}{dt}\frac{d\theta}{dt} + r\frac{d^2\theta}{dt^2} \)

Hint:

In polar coordinates, always separate the acceleration into radial and transverse components: \( a_r = \ddot{r} - r\dot{\theta}^2 \), \( a_\theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} \).

Answer 1

The Correct Option is B

Step 1: Express velocity in polar coordinates.
The velocity vector is \[ \vec{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}. \] Step 2: Differentiate to find acceleration.
\[ \vec{a} = \frac{d\vec{v}}{dt} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}. \] Step 3: Identify the radial component.
The radial component is the coefficient of \( \hat{r} \): \[ a_r = \ddot{r} - r\dot{\theta}^2. \] Step 4: Final Answer.
Hence, the radial component of acceleration is \( \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2. \)