Direction of velocity in curvilinear motion

Question

My textbook mentions that the direction of net instantaneous velocity is along the tangent to the curve. Well, from the polar coordinate system I also know that the net instantaneous velocity has two components: one along the radius of curvature and the other perpendicular to it. But when I sum these two velocities the resultant is not along the tangent to the curve. A simple explanation with an elementary knowledge of vectors would highly be appreciated.

Answer 1

from the polar coordinate system I also know that the net instantaneous velocity has two components: one along the radius of curvature and the other perpendicular to it

In this statement, both the first, radial, component of the velocity and the "other", transverse, component of the velocity should refer to the radius of a circle originating at the center of the polar coordinates or to the position vector - not to the radius of the curve.

If you take that into consideration, there should be no contradiction between your two statements.

Answer 2

Consider a particle with path defined by $(x(t),y(t))$. Two dimensions will do.

The velocity vector is then defined as follows.

$$ \frac{d}{dt}\left[ x(t),y(t)) \right] = (v_x(t),v_y(t)) $$

Now, what is a unit tangent vector? It's a vector that is parallel to the curve $(x(t),y(t))$ at the time t. But how do you determine that? It's the same derivative, only it needs to be normalized to unit length.

That is, the instantaneous velocity vector is defined to be tangent to the curve.

BUT! Note that the tangent vector does not have to be perpendicular to the radial line from the origin. If the curve happens to have non-constant radius, then $\frac{d}{dt} r$ will be non-zero. So we get this.

$$\frac{d}{dt} [r^2] = \frac{d}{dt} [(x(t),y(t)) \cdot (x(t),y(t))] = 2 (v_x(t),y_x(t))\cdot(x(t),y(t))\neq 0 $$

In other words, if the radius is changing then the tangent to the curve is not perpendicular to the radial line. If the radius is not changing then the tangent to the curve is perpendicular to the radius.

Answer 3

My textbook mentions that the direction of net instantaneous velocity is along the tangent to the curve.

The curve being referred to is the trajectory of the particle, and not some arbitrary curve. By definition, velocity vector is everywhere tangential to the trajectory.

Well, from the polar coordinate system I also know that the net instantaneous velocity has two components: one along the radius of curvature and the other perpendicular to it. But when I sum these two velocities the resultant is not along the tangent to the curve.

Which curve? Are you referring to a circle, which is the $r=$constant coordinate curve? There is no reason why the velocity vector must be tangential to this particular curve (after all, choice of a coordinate system is arbitrary). Velocity vector is tangential only to the trajectory of the particle/body.

Answer 4

Consider a particle moving along a path (dashed line below).

At an arbitrary time the velocity vector (green) is tangent to the path. The radial position is $r$ and the azimuth angle $\theta$. The velocity vector is decomposed along the radial and hoop directions with components $\dot{r}$ (blue) and $r \dot{\theta}$ (purple) as seen above.

$$ \boldsymbol{v} = v \hat{\boldsymbol{t}} = (\dot{r}) \hat{ \boldsymbol{r}} + (r \dot{\theta}) \hat{\boldsymbol{\theta}} $$

Moreover, the change in the velocity components (acceleration vector) yield the following four components. Change in radial speed has two components (red) in the radial $\ddot{r}$ and the hoop $\dot{r}\dot{\theta}$. Change in hoop speed has two components also (yellow) with $-r \dot{\theta}^2$ in the radial direction and $\dot{r}\dot{\theta} + r \ddot{\theta}$ in the hoop direction

$$ \boldsymbol{a} = \dot{v} \hat{\boldsymbol{t}} + \frac{v^2}{\rho} \hat{\boldsymbol{n}} = (\ddot{r} - r \dot{\theta}^2) \hat{ \boldsymbol{r}} + (r \ddot{\theta} + 2 \dot{r}\dot{\theta}) \hat{\boldsymbol{\theta}} $$

Above $\hat{ \boldsymbol{t}}$ is the tangential direction (to the path), $\hat{\boldsymbol{n}}$ the normal direction (to the path), $\hat{\boldsymbol{r}}$ the radial direction and $\hat{\boldsymbol{\theta}}$ the hoop direction.