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Ball launched throught semicircular chute

  1. Sep 19, 2007 #1
    I don't get it!!

    problem: A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2g. And the distance between the bottom and the top of the chute is 2R.

    How far from the bottom of the chute does the ball land?

    I realize it is just another projectile motion prob. I calculated the velocity to be (2gR)^(1/2) using the formula (m=v2/radius) for circular motion. and the time was derived using kinematic equation (t=(2R/g)^(1/2)). I DON'T KNOW WHAT I AM DOING WRONG. Sorry, but I'm so frustrated. Is there anyone out there who can solve this?????
  2. jcsd
  3. Sep 19, 2007 #2
    I really couldn't visualize the question, so I can't solve this.

    But if you know the answer, try 3gR or gR instead of 2gR
    or changing values a bit(like plugging different values, like 3gR for 2gR), and when you find the answer, with correct values..then just try to find the meaning... that's what I do when I don't get right answers.
  4. Sep 19, 2007 #3


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    The time to fall a distance 2R at an acceleration of g satisfies 2R=(1/2)gt^2 so I get t=sqrt(4R/g). The acceleration of gravity is still g, not 2g.
  5. Sep 19, 2007 #4
  6. Sep 19, 2007 #5
    2.828R ... is that the answer? i think dick, that you are right about the time being t=sqrt(4R/g) . but my calculation for velocity is correct right? SO if the two values of time and velocity are correct and i multiply them together, i will get 2.828R. How does this look guys???
  7. Sep 19, 2007 #6
    oh. the formula for circular motion is (a=v2/radius) not (m=v2/radius)!
  8. Sep 19, 2007 #7


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    The acceleration of gravity isn't 2g, the magnitude of the centripetal acceleration is 2g.
  9. Sep 19, 2007 #8
    yea, i also 2Rsqrt(2)

    "the ball has a centripetal acceleration of magnitude 2g."

    but you wrote it urself that it's a[c]
    Last edited: Sep 19, 2007
  10. Sep 19, 2007 #9


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    If that's sqrt(8)*R, yes.
  11. Sep 19, 2007 #10
    yes, thanks a lot for clearing that up ^ I have taken that into consideration and reworked the prob. i just need to know if i was successful or if i was still having problems: anyone know if the answer is 2.828R??
  12. Sep 19, 2007 #11
    Thanks! You guys are too cool :)
  13. Sep 19, 2007 #12
    try to get perfect at that technique (I decribed in #2) lol
    It saves a lot of time, and always helps when you need some preferential answer(like in lab's experiments)!
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