Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with projectiles

  1. Nov 12, 2003 #1
    Ok so i need a tinsy bit of help with a lab. It isn't technically hw, so i didn't post in the hw help, plus its not really like a problem per se either.

    So what we need to do was create a device that would make a projectile. We just made a ramp shaped like a V and a marble that slides down it and shoots off. The ramp levels off so there's no Y component to the velocity. Its a nice device, reproducable results and all that. What we need to do is know the marbles flight path so that our teacher can set up a ring at any length along its path, and we'll calulate the height so that the marble will fly through. The ring will be level with the ground.

    here's the part i need a little help with, or i want to double check that what i'm doing is right. We'll do some test runs and use the times and distances it produces to find the X velocity initial of the marble. We alreasy know that the Y velocity initial is zero right? Y acceleration is 32 ft/s2 yes? no acceleration in the X direction once it leaves the ramp yes?

    So once we have velocity in the X we'll have all the info we need right? Then when he gives us the distance we need to go, we'll find how long it takes the marble to travel that distance, and use it in the equation to find how far in the Y direction it goes,(the height we'll need to place the ring.)

    am i wrong with anything? something i'm missing? will everything be alright and accurate and ok?

    thanks yall, the help is much appreciated
     
  2. jcsd
  3. Nov 12, 2003 #2

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You are correct that the horizontal and vertical motions are independent. Gravity affects only the vertical motion.

    Your teacher is presumably going to give you either a horizontal or a vertical distance, and have you calculate the other.

    If your teacher says "the ring should a vertical distance y below the end of the ramp:" First calculate the time it will take for the projectile to fall that distance, considering that the original vertical speed is zero, and the vertical acceleration is g. The horizontal velocity is constant during the time the projectile is falling, so you can caclulate exactly how far it'll fall horizontally during the time it takes to fall a distance y vertically.

    If your teacher says "the ring should be a horizontal distance x from the end of the ramp:" First calculate the time it will take for the projectile to travel that horizontal distance, assuming a constant horizontal velocity at all times. Then figure out how far the projectile will fall vertically in that time, considering the original vertical velocity is zero, and the vertical acceleration is g.

    Does this make sense?

    - Warren
     
  4. Nov 12, 2003 #3
    yeah warren, that all sounds ok. But how do i get the horizontal velocity? test runs like i said? those won't be exactly accurate. Is there another way to calculate V initial in the horizontal direction?
     
    Last edited: Nov 12, 2003
  5. Nov 12, 2003 #4

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The conservation of energy. The mass loses some potential energy as it falls, and gains the same amount of kinetic energy. You might say that, in falling, the mass trades potential for kinetic energy.

    The potential energy lost is (m g y). The kinetic energy of a mass moving at velocity v is (1/2 m v2). Since the amount of poential energy lost equals the amount of kinetic energy gained,

    m g y = 1/2 m v2

    Solve for v:

    v = sqrt(2 g y)

    The velocity of the projectile at the bottom of the ramp has nothing to do with its mass, or the shape of the ramp -- all that matters is the total vertical distance the mass descends.

    Since you're rolling a marble down a ramp, I should tell you that not all of the potential energy lost goes into kinetic energy. A bit of it is lost to friction. A bit more ends up being stored in the rotational inertia of the rolling ball. (Simple analogy: a spinning fan blade has rotational inertia, and can hurt you if you touch it -- energy is "stored" in the rotation.) By and large, however, these two effects (friction and rotational inertia) are negligible.

    - Warren
     
  6. Nov 12, 2003 #5
    alright i have no idea what that stuff means. But all that will give me the horizontal component of the velocity when the marble leaves the ramp??
     
  7. Nov 12, 2003 #6
    Perhaps you haven't studied conservation of energy yet ... yes, chroot's formula is correct under the assumptions he stated.

    However, if you are expected to figure this out yourself and haven't learned conservation of energy, there are other ways to calculate it.

    One way is, as you said, to do a bunch of test runs. No, it won't be exactly accurate --- but neither will a formula! This is the real world, and even if you can ignore friction and rotational inertia and such, there is always random error; slight differences in how the ball is released, how it rolls, etc. So if you're allowed to do test runs, it's a perfectly legitimate way of finding the velocity you need to make further predictions.

    Another way is to just work out all the forces. If you want to ignore friction and rotational inertia, pretend it's a block sliding down an inclined plane. (Your description of the ramp as a "V" shape suggests that the inclined part of the ramp is a plane.) Without the ramp, the block would fall with an acceleration a=g, but with the ramp, the component of the acceleration in the direction down the ramp (as opposed to straight down) is a=g sin(θ), where θ is the angle of the ramp. If you assume the length down the ramp is L (so the height of the ramp is h = L sin(θ)), you can use kinematics to derive chroot's formula without appealing to conservation of energy.
     
  8. Nov 13, 2003 #7
    hmm well.... we did it by test runs. it was accurate enough. The ring we had to get out ball through was barely bigger than our ball so we had to be pretty exact. And our math and physics (which i was in charge of) was right on. We didn't get it through the ring though... What happened was that the when they placed the ring the legnth away i said and the height i said, they didn't put it on the line that the ball taveled on... so it bounced off the side of the ring. It wasn't so bad though, our grade won't be terribly hurt, but now we'll know to be more careful next time. This was our first lab.

    Now i have to write up a lab report and its supposed to be like with html code and on the computer and all that... stuff i'm not exactly expert with....
     
  9. Nov 13, 2003 #8

    turin

    User Avatar
    Homework Helper

    Are you sure about this? I calculated that the rotational kinetic energy would encompass about 29% of the total kinetic energy (effectively slowing the marble down quite a bit).

    whoops! sorry turin!
     
    Last edited by a moderator: Nov 13, 2003
  10. Nov 13, 2003 #9
    Yes, the final speed would be √(10/7 / 2); or about 85% of what is predicted without rotational inertia. Whether that's "negligible" depends on the tolerances of the lab.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with projectiles
  1. Projectile help (Replies: 3)

Loading...