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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.

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    $\begingroup$ There is no component of velocity along the radius of curvature. Probably you are conflating with acceleration. $\endgroup$ – Sayan Mandal Jul 6 '18 at 19:44
  • $\begingroup$ @SayanMandal : There certainly is a radial component to velocity because radius changes in this case. $\endgroup$ – Global Jul 6 '18 at 19:50
  • $\begingroup$ @sayan mandal yes i am talking about variable radius and dr/dt is the radial component of velocity in such cases $\endgroup$ – chemophilic Jul 6 '18 at 20:07
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    $\begingroup$ @MohdKhan - From the description you provided, I agree with Sayan: There is no component of velocity "along the radius of curvature". If you think that there is a misunderstanding here, it would be best for you to provide a diagram in order to clarify your question. $\endgroup$ – user93237 Jul 6 '18 at 20:12
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    $\begingroup$ @MohdKhan I think what you mean is that in the polar co-ordinate system there is a component along the radius vector (not the radius of curvature) and another component along the transverse vector. You should edit your question to remove the words of curvature. ... You should also provide some proof of your claim that these 2 components do not add (as vectors) to the tangential velocity along the curve, which is not true. $\endgroup$ – sammy gerbil Jul 7 '18 at 14:17
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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.

enter image description here

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

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  • $\begingroup$ will you plz add a diagram illustrating the same $\endgroup$ – chemophilic Jul 7 '18 at 5:34
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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.

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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.

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

CurvedPath1

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.

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