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Imagine an object falling along a quarter circle of height $h$ meters ($h$ would also be the radius) and gravity $g$ meters per second squared. How long will it take for the object to travel the distance of the curve in terms of $h$ and $g$? (no friction, no rolling, the object is negligibly small compared to the quarter circle, and the object is stationary until $t = 0$.)

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I was able to determine the following answer:

$${t(h, g)=\sqrt{\frac{h}{g}}\cdot1.854074...}$$

The only issue is that I have no idea where the magic 1.854074... number comes from. This was solved with brute force calculation and curve fitting, but I'm sure a proper solution to this problem will better explain the value.

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3 Answers 3

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As John Forkosh said in the comments this can be done in a similar way to the brachistochrone problem problem. But in this case I think this overly complicates things. Instead we can use the fact that we are on a circle to find the speed. From conservation of energy we have: $$\frac{1}{2}mv^2=mgy$$ where $y$ is measured from the top of the circle downwards. Putting this in terms of angular velocity: $$\frac{1}{2}mh^2 \dot \theta^2=mgh \sin(\theta)$$ $$\int \frac{1}{\sqrt{2\sin(\theta)}}d\theta=\sqrt{\frac{g}{h}}\int dt$$

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  • $\begingroup$ I've looked at the brachistochrone problem and the setup/integration seems complicated and difficult to translate to this problem. Also as a heads up, this is NOT a homework problem. That tag was added by someone else. This was just a curiosity of mine. $\endgroup$
    – Dan
    Commented Aug 13, 2017 at 5:09
  • $\begingroup$ Hey @Dan, what aspects are you having trouble with? Also the 'homework-and-exercise' tag is not just for things that have been set as a homework but for problems you find in e.g. text books instead of conceptual questions. $\endgroup$
    – Quantum spaghettification
    Commented Aug 13, 2017 at 5:13
  • $\begingroup$ I don't know what delta variables mean outside of fractions. Like ds/dt=v makes sense to me, but working with ds or dt outside of that is something I don't understand at all $\endgroup$
    – Dan
    Commented Aug 13, 2017 at 5:18
  • $\begingroup$ @Dan see my edit. $\endgroup$
    – Quantum spaghettification
    Commented Aug 13, 2017 at 5:27
  • $\begingroup$ I'll assume the limits of that first integral should be 0 to pi/2 and sure enough it gives 1.854074..., exactly what I was looking for. Thanks! $\endgroup$
    – Dan
    Commented Aug 13, 2017 at 5:35
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My answer is quite late, but I hope someone will find it helpful.
In fact, we can use Newtonian mech to solve this problem.
We first decompose the gravitational force into two components--tangential and centripetal. The centripetal force does not change the magnitude of the velocity, so let's ignore it. Now we look at the tangential component of the gravity. $$ma_t=mg\cdot\cos\theta$$ we know that $a_t=R\ddot\theta$, so $$R\ddot\theta=g\cos\theta$$ $$\ddot\theta=\frac g l \cos\theta$$ (The equation above is similar to the equation of a SHM pendulum)
Now what we need to do is to solve this ODE. Since this is not easy to do, I will leave the answer like that.

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I am generally averse to assist with anything resembling a homework problem but, in this instance, I think answering would be appropriate.

The question being asked is technically a wholly other question in disguise: the information presented defines a pendulum and you are being asked to find HALF the period.

The formula to find the complete period is: $$T ≈ 2\pi \sqrt\frac{L}{g}$$

Just remember that, in this instance, L and h are the same and you only want HALF the complete period.

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  • $\begingroup$ The formula $T\approx 2\pi \sqrt{L/g}$ is only valid in the case of small oscillations. This is far from true here - given that our 'pendulum' goes to the horizontal. $\endgroup$
    – Quantum spaghettification
    Commented Aug 13, 2017 at 5:29

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