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Consider a pendulum doing vertical circular motion and having velocity with magnitude $v$, mass $m$ and length of string $l$ starting from horizontal (inextensible rope, the given velocity is at horizontal position).

Using force balance at top most point (and considering tension is 0),

$$ \frac{mv_{tngt}^2}{r} = mg$$

Which gives, $$ |v_{tngt}| = \sqrt{rg}.$$

Now by the consideration that at top most point the vertical velocity must turn to zero,

$$ v_{tngt} = |v_{tngt}| \vec{i}.$$

Now, consider another way of doing this using constraint equations, the pendulum moves in a path of circle centered at origin:

$$ x^2 +y^2 = l^2.$$

Differentiating this constraint equation with respect to time,

$$ x v_x + y v_y = 0.$$

Now, this equation relates the components of velocity in cartesian coordinates but the problem is that if you put the $ (x,y)$ of top most point which is $ (0,l)$ there is no useful information to be derive:

$$ 0 v_x + l v_y = 0 .$$

And similarly this happens for the point $ (l,0) , (0,-l) , (-l,0)$ at these points we can't derive any useful information of motion about motion using constraint. I can't understand why at at these points these equations do not give us new information.. as in , is there a deeper understanding to it than it being an algebraic result?


After some deep thought, I noted that all these points are where the branch cuts of the circle equation occurs if we write the constraint as an explicit function of $x$ or $ y$.

For example, if we wrote the isolate the circle equation for $x$ and square rooted, we would get two equations,

$$ x = \sqrt{r^2 - y^2}$$

$$ x= - \sqrt{r^2 -y^2}.$$

This two equations we got from square root can be thought of the circle being split into two half discs along the vertical line $ x=0$ with each equation of the above denoting one half disc. This vertical line cuts the circle at $ (0,l) $ and $ (0,-l)$ so I am guessing the resolution has to do with something about where the explicit function's definition shifts but I'm not sure.

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Why do you say you lose information? $\ell v_y=0$ tells you that the motion is horizontal (i.e. it is consistent with the constraint you’ve imposed), which is all the information you could hope to get from the constraint alone.

The only reason your first calculation contains information about the speed is that you have added in the extra information that the tension is zero by hand. There’s no reason this should be true in general - the vertical component of the force from the rod could be positive, negative, or zero depending on how fast the bob is moving at the top of the arc.

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  • $\begingroup$ Oh I got confused between information brought up from physics arguments and mathematical ones. Thank you! $\endgroup$
    – Brian
    Commented Sep 26, 2020 at 13:20

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