So I was doing an exercise and they find find the angle between the tangential of trajectory and the vertical

I was stuck but then I read in a site that the velocity is ALWAYS tangential to the trajectory

I think this is wrong because in polar v = dr/dr (r) + r d(theta)dt (theta) and the position vector is r = r (r) clearly if you do a scalar (r) . (v) = r (dr/dr ) (r) and not 0 so that means the velocity isn't always tangential?

  • $\begingroup$ possible duplicate of Normal force of ball sliding on concave surface $\endgroup$ – ja72 Dec 8 '14 at 3:15
  • $\begingroup$ a) Please try to use math formatting for better readability and b) In polar coordinates the unit direction vectors do no line up with the path and hence both components are non-zero. Only when rotated along the path the velocity component is zero normal to the path direction. $\endgroup$ – ja72 Dec 8 '14 at 3:21

Careful between $\hat{r}$ and $\vec{r}$. The trajectory $\vec{r}(t)$ is a function that maps $t$ to a position vector $\vec{r}$. But the tangent to the trajectory is not the same as $\vec{r}$. Thus, $\vec{v}\cdot \vec{r}$ tells you nothing about the tangent vector.

To be tangent to something means to be going in the same direction at exactly one point. The line $\vec{r}_0 + \vec{v}\; (t-t_0)$ touches the trajectory $r(t)$ at the point $\vec{r}_0$ at time $t_0$. Moreover, if we take $\vec{v} = \frac{d\vec{r}(t_0)}{dt}$, it is in the same direction, because at $t_0$, the object/whatever is moving in the pointed to by $\vec{v}$. This is almost true by definition of what a tangent vector is.

  • $\begingroup$ But, how do you represent the trajectory? Leave polar coordinates, let's speak in Cartesian coordinates! A trajectory can be represented as x(t), y(t), v(t), or as y(x), z(x), or other ways. Velocity is dr/dt, (r = vector), but the answer to the question may vary according to the representation of the trajectory. $\endgroup$ – Sofia Dec 7 '14 at 14:54

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