Physics olympiad problem -- struggling with polar coordinates

[cr7einstein]

@ehild thanks a lot; I have understood it now...thank you for your patience and understanding.

 

[ehild]

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The particle acceleration is the time derivative of the velocity vector, and, using what you have been learning for polar coordinates, is given by:
$$\vec{a}=\frac{d\vec v}{dt}$$
From a force balance on the particle, we know that the tangential component of acceleration is equal to zero (frictionless track), right? The tangential component of acceleration is equal to the acceleration vector dotted with a unit tangent vector to the trajectory. But the velocity vector is tangent to the trajectory. So, the unit vector tangent to the trajectory is equal to the velocity vector divided by its own magnitude. Since the dot product of the acceleration and a unit tangent vector to the trajectory must be equal to zero, the dot product of the acceleration vector with the velocity vector must also be equal to zero.

So, if we take the dot product of the acceleration vector with the velocity vector we obtain zero, for any horizontal frictionless track.
That means the time derivative of the velocity is perpendicular to the velocity. But $\frac{d(\vec v)^2}{dt}=2\vec v \dot{ \vec v} = 0$ as they are perpendicular. That means the time derivative of v² is constant, the speed is constant.

 

[Chestermiller]

So, if we take the dot product of the acceleration vector with the velocity vector we obtain zero, for any horizontal frictionless track.
That means the time derivative of the velocity is perpendicular to the velocity. But $\frac{d(\vec v)^2}{dt}=2\vec v \dot{ \vec v} = 0$ as they are perpendicular. That means the time derivative of v² is constant, the speed is constant.

Yes. That's what Haruspex was hinting at in post #24, but apparently, it did not resonate with the OP. It's not surprising, considering that the OP was considering $\theta$ to be an independent variable rather than the dependent variable (see post #34).