cancel
Showing results for 
Search instead for 
Did you mean: 
cancel
Showing results for 
Search instead for 
Did you mean: 

Community Tip - Help us improve the PTC Community by taking this short Community Survey! X

Mathcad Community Challenge May 2023 - Optimize Trajectory for Maximum Horizontal Distance

DaveMartin
16-Pearl

Mathcad Community Challenge May 2023 - Optimize Trajectory for Maximum Horizontal Distance

A ball with a mass of 1 kilogram is at the top of a frictionless ramp 10 meters above the ground. The ball rolls down the incline and launches from a height of 2 meters and an angle of 30 degrees above the ground.

DaveMartin_0-1682971863452.png

 

(This picture was created in Creo.)

  1. Create a function that calculates the horizontal distance as a function of initial height, launch height, and launch angle.
  2. Calculate the horizontal distance the ball will land from the end of the ramp.
  3. Solve for the angle that will optimize the horizontal distance.
  4. How will the horizontal distance change if this were performed on the Moon instead of on the Earth’s surface? Assume the acceleration due to gravity on the Moon’s surface is 1/6 that of Earth.
  5. Use the Chart Component to depict how the horizontal landing distance changes as a function of angle.
  6. Use a 3D Plot to show how the horizontal landing distance changes as a function of ramp height and launch angle. Assume the ball starts at a height of 10 meters.

The worksheet should contain sufficient documentation to stand on its own; someone unfamiliar with the initial problem should be able to understand what is being calculated.

Find the Mathcad Community Challenge Guidelines here!

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com
33 REPLIES 33

Can we assume that the ramp angle (from 10 meters) is also 30 degrees?  (Otherwise the horizontal velocity requested in #1 is not fixed.  Or do you want that discussed too?

The function should calculate horizontal distance, not velocity.

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com


@DaveMartin wrote:

The function should calculate horizontal distance, not velocity.


Yes, that is clear, but the distance will probably depend on the speed reached.
And this still does not answer the question whether the two legs of the ramp actually enclose a right angle, as one could assume on the basis of the drawing.

So is the angle of the longer ramp 60 degree to the horizontal plane or should it be considered variable?

Here are a couple hints: potential energy gets converted to kinetic energy. The angle between the two legs of the ramp does not matter. I almost threw in the horizontal length of the ramp as a red herring, but I thought that might get people too caught up in the angles of the ramp. And even though I drew the angle between the two legs of the ramp at 90 degrees, it actually doesn't factor into the answer.

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com

OK, but when the ball reaches zero height and is supposed to switch on to the 30° ramp, a part of the energy would be transformed into deformation and heat (even when you assume frictionless motion). How much depends on the angle of the first longer ramp.

But I assume that you want us to neglect this as well and also assume a non-elastic ball with a diameter of zero so also the geometry where the two ramps meet is not of importance, right?

Yes, please ignore deformation, heat, and diameter/volume of the ball. Essentially it's a trajectory problem where the initial velocity comes from potential energy.

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com

My Attempt

ppal_0-1683081918937.pngppal_1-1683081943373.png

 

ppal
17-Peridot
(To:DaveMartin)

So i get 

ppal_0-1683072742540.png

for the first iteration . Used Moment of inertia of hollow sphere .

 

Anyone can confirm or dissent before I go further?

Werner_E
24-Ruby V
(To:ppal)

I have not spent much time on it so I may had made mistakes,  but I ended up at approx, 17 meters

Werner_E_1-1683079355241.png

 

The desired function would be

Werner_E_0-1683085448658.png

 

or

Werner_E_1-1683085478039.png

 

 

 

 

ppal
17-Peridot
(To:DaveMartin)

Comments on my approach - Prime 9 attached

 

 

Werner_E
24-Ruby V
(To:ppal)

Since the size of the sphere is not specified, I assume that it is simplified to be a point mass and things like rotational energy should not matter. @DaveMartin  could go into more detail about this.

EDIT: Ahh, I just noticed that he already wrote that the diameter of the ball should be ignored as well. So I guess that we should also assume that all kinetic energy is in translation and none in rotation. The ball is sliding down and up the ramp and not rolling, hmm...

 

Yes, please consider this a point mass. This problem was inspired by a high school physics problem cited by NASA in the 1960s. My intent was for people to consider the potential energy for the object at the top of the ramp and convert that into the kinetic energy as it comes off the ramp. That will give you the velocity of the object, and then you can plug that into trajectory equations like:

https://www.physicsgoeasy.com/maximum-range-of-projectile-formula/?utm_content=expand_article

For optimizing the distance in step 3, people should have an intuitive sense that the angle would be 45 degrees. I wanted to see how people would calculate that in Mathcad.

The main point was to get to charts and 3D plots in steps 5 and 6.

I should have probably used the word "slides" instead of "rolls" in the explanation of the problem. I will see if I can edit that. You can ignore friction, air resistance, rotational inertia, heat dissipation, or any other effects. I wasn't looking for anything beyond potential energy going into kinetic energy.

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com


For optimizing the distance in step 3, people should have an intuitive sense that the angle would be 45 degrees.

Yeah, thats also what many may remember from school but it ignores the 2m ramp height above floor

Werner_E_0-1683154549581.png

 

Werner_E_1-1683154638490.png

 


I should have probably used the word "slides" instead of "rolls" in the explanation of the problem.


Would be unusual for a ball to be sliding instead of rolling 😉 So you would have to use a different point object which slides.

But #1 needs clarification. Which "horizontal distance"? Probably you mean the horizontal distance from the end of the launch ramp depending on time (additionally to the asked for initial height, launch height, and launch angle). So a function with four arguments, not just three. But without friction, air resistance, etc. its just a simple function linear in time.
If you meant in #1 a function for the landing distance,  #2 would simply mean to evaluate that function for the given input values.

 

 

Yes, it's not exactly 45 degrees as I altered the original problem. I changed it slightly so that the ground is not level with the end of the ramp, as shown in this problem:

https://homework.study.com/explanation/a-block-slides-down-a-frictionless-incline-that-terminates-in-a-45-degree-ramp-as-shown-in-figure-3-find-an-expression-for-the-horizontal-range-x-shown-in-the-figure-as-a-function-of-the-heights-h1-and-h2-the-solution-for-this-is-the-range-x-2-h1-h.html

Let's please keep it simple. I wasn't going for anything tricky or complex. In #1, I mean the horizontal distance from the end of the ramp when the ball lands.

Yes, a ball rolls, it doesn't slide. But I had a nice model for a ball in Creo that I could use to make the image, so I did.

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com

So anyway here is my entry - out of competition as it was created in Mathcad 15. But this version allows a much quicker and faster workflow and is significantly more convenient to use. Quite apart from the fact that it is also more powerful in terms of functionality.

I briefly tried converting the file to Prime 9, but despite the fact that it would have required a lot of reformatting to make it look anything like the same, I also had to refuseto put up with the many shortcomings of Prime and the converter to bang around. The well known auto-label bug has struck again - h has become a constant, g an undefined variable without a way to change it. And Prime's inability to deal with incomplete functions would have required a workaround in calculating the optimal angle. I didn't even get as far as the plots then. I quickly gave up trying to create a Prime Worksheet, very disgusted, and instead attached a PDF Print in addition to the Mathcad file.

Have fun!

Here the pics of the two pages

Werner_E_3-1683166647115.png

 

Werner_E_2-1683166615025.png

 

 

Looks like you got the trick answer to #4. The acceleration due to gravity falls out of the equation. So it doesn't matter if you're doing it on the Earth, the Moon, or Pluto.

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com

Solution attached, Prime 4.0 Express

JeffH1
14-Alexandrite
(To:DaveMartin)

My submission in Prime 9.0 format.

 

ttokoro
20-Turquoise
(To:DaveMartin)

Ignore rotational kinetic energy. Prime 9. No Chart Component but x-y plot. Without units to solve.

image.png

Here's my limited attempt with Prime 8 Express:

May1.pngMay2.pngMay3.png

If it's a ball, why does the incline need to be frictionless?  The intent of a ball is to have it rotate.  For that to occur, you need nominal friction.  

If you want the incline to have friction, add friction. If you want to incorporate that and rotational inertia, have at it.

I will repeat my original intent. I modified an old high school problem, where the intent was for people to use the potential energy to find the initial velocity to calculate a trajectory. Anything else you want to add to the problem is up to you.

As with the previous Mathcad Challenges, the point isn't the answer you get, but how you choose to approach the problem with the tools in Mathcad.

Dave Martin - dmartin@creowindchill.com - https://www.mcaeconsulting.com
JeffH1
14-Alexandrite
(To:JE_9022802)

@JE_9022802 - I'm with you.  As much as 30% of the kinetic energy goes into rotational inertia.  Without it, the problem is non-physical and sets a bad precedent, even for high schoolers.  It is workable with the rotational component.  Give it a go.

ppal
17-Peridot
(To:DaveMartin)

 

Prime 9 file

ppal_1-1683687908326.png

ppal_3-1683688015259.png

ppal_4-1683688044879.png

ppal_6-1683688080344.png

ppal_7-1683688110821.png

 

ppal_8-1683688124629.png

 

ppal_9-1683688142994.png

 

ppal_10-1683688165569.png

 

ppal_11-1683688180794.png

ppal_12-1683688198229.png

ppal_13-1683688231661.png

 

 

 

 

 

 

Good evening fellow friends of Mathcad Prime,

 

Please find my contribution in attachment (Mathcad Prime 9).

 

Kind regards,

 

Jan Claeys

Looks great. Is your launch height and landing site at the same level? 

Jan_Claeys_9A
6-Contributor
(To:ppal)

You're right: I've made a mistake and I'll correct this tonight...

Jan_Claeys_9A
6-Contributor
(To:ppal)

In attachment the reworked version v2

ppal
17-Peridot
(To:DaveMartin)

Another one in Prime 9. Bit more compact. perhaps harder to follow:

Top Tags