Question

A particle is moving with a constant angular velocity 2 rad/s in an orbit on a plane. The radial distance of the particle from the origin at time t is given by \(r = r_0 e^{2\beta t}\) where \(r_0\) and \(\beta\) are positive constants. The radial component of the acceleration vanishes for \(\beta = \) \rule{1cm{0.15mm} rad/s. (in integer)}

Hint:

It is essential to memorize the expressions for velocity and acceleration in polar coordinates for mechanics problems. Velocity: \(\mathbf{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}\) Acceleration: \(\mathbf{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\theta}\) The term \(-r\dot{\theta}^2\) is the centripetal acceleration, and \(2\dot{r}\dot{\theta}\) is the Coriolis acceleration.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
This problem deals with the kinematics of a particle moving in a plane, described using polar coordinates (\(r, \theta\)). The acceleration of the particle has two components: a radial component (\(a_r\)) and a tangential or angular component (\(a_\theta\)). We need to find the condition under which the radial component is zero.
Step 2: Key Formula or Approach:
In polar coordinates, the acceleration vector is \(\mathbf{a} = a_r \hat{r} + a_\theta \hat{\theta}\). The radial component of the acceleration is given by the formula: \[ a_r = \ddot{r} - r\dot{\theta}^2 \] where \(\dot{r}\) and \(\ddot{r}\) are the first and second time derivatives of the radial distance \(r\), and \(\dot{\theta}\) is the angular velocity \(\omega\).
Step 3: Detailed Explanation:
We are given: Constant angular velocity: \(\dot{\theta} = \omega = 2\) rad/s. Radial distance as a function of time: \(r(t) = r_0 e^{2\beta t}\). First, we need to find the first and second time derivatives of \(r(t)\): First derivative (\(\dot{r}\)): \[ \dot{r} = \frac{d}{dt}(r_0 e^{2\beta t}) = r_0 (2\beta) e^{2\beta t} = 2\beta (r_0 e^{2\beta t}) = 2\beta r \] Second derivative (\(\ddot{r}\)): \[ \ddot{r} = \frac{d}{dt}(2\beta r_0 e^{2\beta t}) = 2\beta r_0 (2\beta) e^{2\beta t} = 4\beta^2 (r_0 e^{2\beta t}) = 4\beta^2 r \] Now, substitute these into the formula for the radial acceleration \(a_r\): \[ a_r = \ddot{r} - r\dot{\theta}^2 = (4\beta^2 r) - r(2)^2 = 4\beta^2 r - 4r \] The problem states that the radial component of the acceleration vanishes, so we set \(a_r = 0\): \[ 4\beta^2 r - 4r = 0 \] Factor out \(4r\): \[ 4r(\beta^2 - 1) = 0 \] Since \(r = r_0 e^{2\beta t}\) is not generally zero (as \(r_0\) is a constant and \(\beta>0\)), the term in the parenthesis must be zero: \[ \beta^2 - 1 = 0 \] \[ \beta^2 = 1 \] \[ \beta = \pm 1 \] The problem states that \(\beta\) is a positive constant, so we choose the positive root. \[ \beta = 1 \] Step 4: Final Answer:
The radial component of the acceleration vanishes for \(\beta = 1\) rad/s.