Time calculation with non uniform acceleration

In summary: It saves energy because the skateboarder is able to use less energy to get to the bottom of the half-pipe.
  • #1
woodie37
14
0
How long does it take a skateboarder to roll from the top of a half pipe of radius 2 meters to the bottom? Ignore friction, and assume the half pipe is in the shape of a semi circle. I tried to use integration to solve this but failed miserably, would someone please show me how to do this? Show full solution please.
 
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  • #3
Welcome to PF!

Hi woodie37! Welcome to PF! :smile:

Use conservation of energy to find angular speed as a function of angle, and then integrate. :wink:

Show us what you get. :smile:
 
  • #4
We love to help students like you, but please show us what you've tried so we can help :).

You can use either integration or conservation of energy as tiny-tim has said.

Using integration is somehow more difficult, and has a longer path, and will lead eventually to the law of conservation of energy.

But using the law of conservation of energy is easier, and will give you a very direct answer.

Good luck :)
 
  • #5
But will he be able to "work" with time in all this calculations??

Would't it be the same than calculating the pendulum equation?

It happened to me that I tried for many months when i was younger (about 16 y.o.) but y could only find w(&), that is to say omega (theta) what is angular speed in function of angle.

But I always wondered if it was possible to aply power relationships to the energy conservation in order to calculate such times. Then, is it possible?

Thanks in advance and sorry my english.
 
  • #6
That's what I've been saying, pal. He can either use the chain rule with circular motion laws, and derive the energy conservation laws, or he can directly apply the energy conservation laws, which I can do in 2 lines!
 
  • #7
I don't think he's going to show up again, by the way. I think he expected us to solve problems for him, that's very bad. Anyway if you want to discuss the answer you're welcome too ;). Show us something and let's talk ;)
 
  • #8
Hi TheDestroyer! :smile:
TheDestroyer said:
… or he can directly apply the energy conservation laws, which I can do in 2 lines!

Conservation of energy saves energy! :wink: :biggrin:
 
  • #9
hehehehehe :D, that's right ;). I really used to do it using chain rules and integrals, and noticed how stupid I had been been when I discovered the energy easy way :P
 
  • #10
but chain rule way is more "mathematically" beautyful :-p
 
  • #11
Yeah, it shows how consistent physics and mathematics are :D, I really was surprised when discovered how whatever you do, if you do it right, it gives you the same answer :D
 
  • #12
Yehh, that's the beauty of natural science even in this very early and simple level.

Good science Destroyer ;)
 
  • #13
Ok sorry about the delay but I haven't been on the computer for awhile, just checked my email and read that someone gave a reply.

My solution is as follows

starting from the top of the half circle, angle x is equal to 0 while after passing a half circle, angle x is 180

Btw I did use conservation of energy to figure out the velocity (not angular) as a function of x

-U=K where U is the potential energy and K is the kinetic
so 2grsinx=v^2 v=[tex]\sqrt{2grsinx}[/tex]
The distance, d = [tex]\pi[/tex]r where r is the radius
and t = d/v = d(2grsinx)^(-1/2)

Then I integrated...

t=[tex]\int[/tex]d(2grsinx)^(-1/2)dx

Then the thing became difficult...someone help me out please
and could someone, while explaining the physics, also give me a quick tutor of how to use the latex reference to insert symbols? because i don't understand programming
 
  • #14
Hi woodie37! :smile:

(have a theta: θ and an omega: ω and a pi: π and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
woodie37 said:
… so 2grsinx=v^2 v=[tex]\sqrt{2grsinx}[/tex]
The distance, d = [tex]\pi[/tex]r where r is the radius
and t = d/v = d(2grsinx)^(-1/2)

Nooo :redface: … use v = r(dx/dt)

(or preferably write θ instead of x, and ω instead of dx/dt, to give you v = √(2grsinθ) = ωr)

[noparse]For LaTeX on PF, just type [tex] before and [/tex] after, [/noparse]and see http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000" for symbols. :smile:
 
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  • #15
ok thanks and if i get more problems i'll ask again =D
 
  • #16
Looking at the bastard of an integral I ended up with, along with the wiki page, is there any way to solve this without resorting to stuff like Elliptic functions?

Other than giving us a function for the angular velocity as a function of sine of theta, how does conservation of energy really help here?

Solving for velocities, accelerations and angular positions is straight-forward, but time mucks everything up. :\ I always did suspect there was a reason we used the small angle approximation.
 
  • #17
Sorry to double-post, but this is an interesting question.
I've looked for a solution for the last couple of days, but the only things I've found are with the Elliptic functions and potential wells and whatnot, and that seems a bit out of the scope of this question. Further insight would be greatly appreciated.
 
  • #18
TheDestroyer said:
He can either use the chain rule with circular motion laws, and derive the energy conservation laws, or he can directly apply the energy conservation laws, which I can do in 2 lines!
If anyone thinks they can solve for the time using conservation of energy, I'd love to see it. The only way I know involves elliptical integrals (as RoyalCat found).

If I'm missing something simple, I await enlightenment. (It wouldn't be the first time I missed the obvious!)
 
  • #19
Can someone post a solution here involving elipticals? I'd like to see a solution please, any would be fine, be it elliptical or be it other. =D
 

Related to Time calculation with non uniform acceleration

1. How do you calculate time with non-uniform acceleration?

To calculate time with non-uniform acceleration, you can use the formula t = (v - u)/a, where t is time, v is final velocity, u is initial velocity, and a is acceleration. This formula assumes that acceleration is constant, but for non-uniform acceleration, you can use calculus to calculate the time by taking the integral of the acceleration function over the desired time interval.

2. Can you provide an example of time calculation with non-uniform acceleration?

For example, if a car starts from rest and accelerates at a rate of 2 m/s^2, then after 3 seconds, its velocity will be 6 m/s. However, if the acceleration is non-uniform, such as increasing at a rate of 2 m/s^2 every second, the final velocity after 3 seconds will not be 6 m/s. In this case, you would need to use the integral of the acceleration function to calculate the time.

3. What is the difference between uniform and non-uniform acceleration?

Uniform acceleration is when the rate of change of velocity is constant, while non-uniform acceleration is when the rate of change of velocity is not constant. This means that for uniform acceleration, the change in velocity per unit of time will be the same, while for non-uniform acceleration, the change in velocity per unit of time will vary.

4. How does non-uniform acceleration affect time calculations?

Non-uniform acceleration can make time calculations more complex because the acceleration is changing over time. This means that the initial formulas for calculating time may not be accurate and you may need to use more advanced mathematical methods, such as calculus, to determine the time accurately.

5. Can time calculations with non-uniform acceleration be applied to real-life situations?

Yes, time calculations with non-uniform acceleration can be applied to real-life situations. In fact, it is commonly used in fields such as physics and engineering to model and analyze the motion of objects. For example, calculating the time it takes for a rocket to reach a certain velocity or the time it takes for a car to come to a stop with varying levels of acceleration.

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