- #1
davidbenari
- 466
- 18
Homework Statement
A particle moves with velocity ##v## along the cardioid ##r=k(1+cos\theta)##. Find ##\vec{r''}\cdot \hat{r}## and interpret the result. At what angles will this coincide with the centripetal acceleration?
Homework Equations
The Attempt at a Solution
To find ##\vec{r''}\cdot \hat{r}## I will simply use the formula for the radial acceleration in cylindrical coordinates. This is ##\rho''-\rho\theta'^2## which after lengthy manipulation gives: ##-k(\theta'' sin\theta + \theta'^2cos\theta)-k\theta'^2(1+cos\theta)##. Which I have no idea how to "interpret", haha.
To solve the second part well I know the position vector is ##\vec{r}=r\hat{\rho}## and that its derivative with respect to ##\theta## will give a tangent vector to the curve. This is simply ##\frac{d\vec{r}}{d\theta}=\frac{dr}{d\theta}\hat{\rho}+r\hat{\theta}##
The unit tangent vector is simply: ##\hat{t}=\frac{1}{\sqrt{(\frac{dr}{d\theta})^2+r^2}}(\frac{dr}{d\theta}\hat{\rho}+r\hat{\theta})##. Which helps me to find the orthogonal normal vector and thus the centripetal acceleration as ##\frac{v^2}{R}\frac{1}{\sqrt{(\frac{dr}{d\theta})^2+r^2}}(\frac{dr}{d\theta}\hat{\theta}-r\hat{\rho})##. Note I've switched components on the and sign on one component as well.
This helps me notice that the centripetal acceleration coincides with my above result when the centripetal acceleration is purely in direction of ##\rho##. This happens when ##\frac{dr}{d\theta}=k(-\sin\theta)=0## which is for Pi and 0.
Now, help me interpret the acceleration expression I derived above to which I'm expected to provide an interpretation. Also, is my procedure correct?
Thanks for reading this tedious thing.