How Long Does a Ball Take to Climb a Parabolic Ramp?

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Homework Help Overview

The discussion revolves around a physics problem involving a ball climbing a parabolic ramp defined by the equation y=(x^2)/3, with an initial velocity of 5 m/s and the influence of gravity. The original poster seeks to determine the time taken for the ball to reach a height of y=1, while ensuring energy conservation throughout the motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of energy conservation principles and differential equations to approach the problem. Some question the necessity of trigonometric functions in their calculations. Others express uncertainty about the problem's complexity and whether it aligns with high school-level knowledge.

Discussion Status

There is an ongoing exploration of various methods to solve the problem, with some participants sharing their attempts and expressing frustration over the complexity of the resulting equations. Guidance has been offered regarding the use of definite integrals and references to similar problems, but no consensus has been reached on a clear solution path.

Contextual Notes

Participants note that the problem is not part of a formal homework assignment but rather an extracurricular challenge. There is a shared concern about the appropriateness of the problem for high school students, with some participants indicating they are still learning relevant mathematical concepts.

errwrsysalan
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1. Assume there is gravity and no external force acting on the system. A ball has an initial velocity of 5 m/s and climbs up a parabolic ramp, which is defined by y=(x^2)/3. If the ball rolls exactly along the path of ramp and energy of the ball is conserved, starting from (0,0), calculate the time taken for the ball to climb to a height of y=1.

2. Ek+Ep=Einit, Vy=||v||sin(arctan(dy/dx))

3. I made a differential equations using the formula above but the differential I got is only numerical integratable (results given by wolfram alpha). I'm looking for an easier way to solve this problem. So, guys, please help! Thanks!
 
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What did you need the trig for?
Did you start with ##\frac{1}{2}m(\dot x^2+\dot y^2) + mgy = \frac{1}{2}mu^2 : u=5,\; y=\frac{1}{3}x^2## ?
 
Simon Bridge said:
What did you need the trig for?
Did you start with ##\frac{1}{2}m(\dot x^2+\dot y^2) + mgy = \frac{1}{2}mu^2 : u=5,\; y=\frac{1}{3}x^2## ?
That seems pretty helpful! I haven't learned to use and x and y things because I'm a high school student, but I've done stage 1 college maths so I think I can cope with that. I'll give it a try! Thank you!
 
Is this a problem set for high school?
 
Well. It's not, I made this question up. It'd be nice if you can solve it using high school knowledge only though.
 
Simon Bridge said:
Is this a problem set for high school?

I'm stuck again...
 
Simon Bridge said:
Is this a problem set for high school?

I might got it.. trying now
 
Simon Bridge said:
Is this a problem set for high school?

I'm trying now, but still, can you show me how you would do it please?

Guess what. I used the formula you gave me and that ends up the same differential equation I got last time using trig, which I need to use a computer to calculate it. Can you help me?
 
Please answer the question?
 
  • #10
Simon Bridge said:
Please answer the question?
errwrsysalan said:
Well. It's not, I made this question up. It'd be nice if you can solve it using high school knowledge only though.

I answered above. Basically this is some extracurricular work.
 
  • #11
Sorry - misread.
 
  • #12
Simon Bridge said:
Sorry - misread.
It's ok. So, right now I used your formula, and it ended up exactly the same as last time I did it using trig.
I got dt/dy=0.5sqrt((3+4y)/(25y-2gy^2)) dy, which even wolfram alpha couldn't give an exact solution.
I believe there is a better way to do this. Can you show me please?
 
  • #13
... this is a specific case of an object sliding inside a parabolic bowl.
If you google for that, you'll get a bunch of examples. i.e. See attachment below.

Note: ##\dot x = v(x)## is where you get stuck.
You can do $$T = \int_0^{\sqrt 3} \frac{dx}{v}$$ ... i.e. what you were doing but as a definite integral.
The indefinite integral involves elliptical functions.

It is not unusual for quite simple-seeming setups to be very difficult to solve - or even have no analytic solutions at all.
ie. Try working out the exact equation of motion for a simple pendulum - or a puck sliding in a spherical bowl.
 

Attachments

  • #14
Simon Bridge said:
... this is a specific case of an object sliding inside a parabolic bowl.
If you google for that, you'll get a bunch of examples. i.e. See attachment below.

I'd usually reach for lagrangian mechanics.
It is not unusual for quite simple-seeming setups to be very difficult to solve - or even have no analytic solutions at all.
Try working out the exact equation of motion for a simple pendulum for example - or a puck sliding in a spherical bowl.

Thank you! No problem. I need to sleep as well. I'll have a look tomorrow.
 

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