Question

Consider a particle of mass m moving in a plane with a constant radial speed \(\dot{r}\) and a constant angular speed \(\dot{\theta}\). The acceleration of the particle in (r,\(\theta\)) coordinates is

• A. \(2r\dot{\theta}^2\hat{r}-\dot{r}\dot{\theta}\hat{\theta}\)
• B. \(-r\dot{\theta}^2\hat{r}+2\dot{r}\dot{\theta}\hat{\theta}\)
• C. \(\ddot{r}\hat{r}+r\ddot{\theta}\hat{\theta}\)
• D. \(\ddot{r}\theta\hat{r}+r\ddot{\theta}\hat{\theta}\)

Answer 1

The Correct Option is B

The problem involves determining the acceleration of a particle moving in a plane with constant radial and angular speeds. We can break down the problem using polar coordinates \((r, \theta)\).

In polar coordinates, the position of the particle is given by the radial distance \(r\) and the angle \(\theta\). The velocities in these coordinates are:

• Radial velocity (\(v_r\)): \(\dot{r}\)
• Angular velocity (\(v_\theta\)): \(r \dot{\theta}\)

To find the acceleration, we need to differentiate these velocities:

1. Radial acceleration (\(a_r\)) is given by differentiating \(v_r\): \(\ddot{r} - r\dot{\theta}^2\). Since \(\dot{r}\) is constant, \(\ddot{r} = 0\), thus the radial acceleration is \(-r\dot{\theta}^2\).
2. Transverse (angular) acceleration (\(a_\theta\)) is given by differentiating \(v_\theta\): \(\frac{d}{dt}(r\dot{\theta}) = r\ddot{\theta} + \dot{r}\dot{\theta}\). Again, since \(\dot{\theta}\) is constant, \(\ddot{\theta} = 0\), so the transverse acceleration is \(2\dot{r}\dot{\theta}\).

The total acceleration \(\mathbf{a}\) in polar coordinates is then: \(\mathbf{a} = a_r \hat{r} + a_\theta \hat{\theta} = -r\dot{\theta}^2\hat{r} + 2\dot{r}\dot{\theta}\hat{\theta} \)

Thus, the correct answer is: \(-r\dot{\theta}^2\hat{r} + 2\dot{r}\dot{\theta}\hat{\theta}\), explaining option 2 is correct based on the calculation of radial and transverse components of acceleration.