# Determine which mass is going to reach the bottom first

## Homework Statement

The problem includes some pictures, so I will post the link here: http://faculty.polytechnic.org/cfle...pdfs _Jan_2010/zq-Mult Choice Test 3 csb.pdf

It is question number 20

Ug=mgh
KE=(1/2)mv^2

## The Attempt at a Solution

Since the inclines are frictionless, the time depends on the mass, velocity, and distance travel. However, the mass is not given, so I don't know how to approach the problem. Can somebody please help me out with this?

time would depend on the initial net acceleration. It would help if you draw free body diagram and check the net force acting on body. Would the net force depend on slope of incline???.---------->The correct answer to this question would answer your question.

Caution: They all will have same velocity at the bottom!!!

The equations marked as relevant are indeed very relevant here. Total energy is conserved, hence ## v = \sqrt {2gh} ##. Now, ## v = ds/dt##, hence $$dt = \frac {ds} {\sqrt {2gh} }$$ As soon as you figure out the dependency ## ds = f(h) dh ##, you will be able to set up and compute the integral $$T = \int\limits_R^0 \frac {f(h) dh} {\sqrt {2gh}}$$

mfb
Mentor
The time does not depend on the mass. Forces will be proportional to mass, but as F=ma, the mass cancels out again when you calculate an acceleration.

[Edit: Merged replies and deleted the second thread, I did not see this thread]

@voko: I doubt that he is supposed to perform that integral for the curved shape in a multiple choice test.

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@voko: I doubt that he is supposed to perform that integral for the curved shape in a multiple choice test.

Hmm. I may be overthinking the problem, but I don't see an easier way to compare #3 with the other two.

mfb
Mentor
The length of the tracks in 2 and 3 is the same, this makes a comparison easy. And 2 and 1 can be compared with each other as well.

The length of the tracks in 2 and 3 is the same, this makes a comparison easy.

I am still not seeing that.

The equations marked as relevant are indeed very relevant here. Total energy is conserved, hence ## v = \sqrt {2gh} ##. Now, ## v = ds/dt##, hence $$dt = \frac {ds} {\sqrt {2gh} }$$ As soon as you figure out the dependency ## ds = f(h) dh ##, you will be able to set up and compute the integral $$T = \int\limits_R^0 \frac {f(h) dh} {\sqrt {2gh}}$$

The final equation has an error, it should be $$T = \int\limits_0^R \frac {f(h) dh} {\sqrt {2gh}}$$ Now, taking ## x = \sqrt {R^2 - h^2} ##, we get ## ds = R (R^2 - h^2)^{-1/2} dh ##. Further taking ## h = R u ##, we end up with $$T = \sqrt {\frac {R} {2g}} \int\limits_0^1 (u(1 - u^2))^{-\frac 1 2} du$$ That is an elliptic integral, so it is very dubious indeed that this is supposed to be the way to solve the problem.

nasu
Gold Member
I don't think that they ask to calculate the time.
Just look at the pictures. The one "closest to a free fall" will be first.

D H
Staff Emeritus
we end up with $$T = \sqrt {\frac {R} {2g}} \int\limits_0^1 (u(1 - u^2))^{-\frac 1 2} du$$ That is an elliptic integral, so it is very dubious indeed that this is supposed to be the way to solve the problem.
I took a different approach and ended up with $$T=\sqrt {\frac {R} {2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{\sin\theta}}$$ This also is an elliptic integral. Your result is the same as mine, just expressed differently.

This is less time than is needed to slide down a straight ramp with the same path length. I too find it dubious that either your approach or mine is the one expected to be used to solve the problem.

If we just want to answer the question as to which one hits the ground the fastest, I suggest you create a problem where you have a two cases. See my diagram. Figure 2 shows a ramp 20 units long with a height of 15. Figure 1 shows a broken ramp with total length 20 and a height of 15. Now it is easy to calculate which one will hit the ground first.

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D H
Staff Emeritus
I don't think that they ask to calculate the time.
Just look at the pictures. The one "closest to a free fall" will be first.
What does that mean, though?

Let's add two more objects to slide down. Masses m4 and m5 traverse the same horizontal distance as that traversed by mass m3. Mass m4 slides down a 45 degree ramp, and for mass m5, the curve followed is a segment of an inverted cycloid. Mass m5 will take the least amount of time and mass m4 the most. How does "closest to free fall" distinguish between these three curves?

haruspex
Homework Helper
Gold Member
2020 Award
I don't think that they ask to calculate the time.
Just look at the pictures. The one "closest to a free fall" will be first.
You can do a bit better than just looking at the pictures. Compare the speeds after travelling some given distance s. in the curved path, it will have descended further for any s less than the whole distance.

• 1 person
nasu
Gold Member
What does that mean, though?

Let's add two more objects to slide down. Masses m4 and m5 traverse the same horizontal distance as that traversed by mass m3. Mass m4 slides down a 45 degree ramp, and for mass m5, the curve followed is a segment of an inverted cycloid. Mass m5 will take the least amount of time and mass m4 the most. How does "closest to free fall" distinguish between these three curves?

I did not say you can "guess" for any arbitrary trajectories.
I am sure you can find any number of paths where you need to calculate to find the actual time.
But here they picked nice examples so you can try a guess without calculations.
The circular path starts with the highest acceleration (g at the top). And even though at some point the acceleration becomes lower than on any of the two inclines, the average speed will be higher and the distance shorter. Of course, this is not a mathematical proof, just some semi-intuitive estimate. But do you really think they expected to solve numerically these integrals?

D H
Staff Emeritus
But do you really think they expected to solve numerically these integrals?
Heavens no! These questions might well be for a non-calculus based physics class.

haruspex' method works quite nicely and is at an appropriate right level.

nasu
Gold Member
You can do a bit better than just looking at the pictures. Compare the speeds after travelling some given distance s. in the curved path, it will have descended further for any s less than the whole distance.

I did not mean look without thinking. Bit I did not explain as well as you did, you are right.