1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Loop the Loop

  1. Apr 1, 2007 #1
    A small spherical ball of radius r = 1.7 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 3.1 m. The ball has mass M = 345 g.

    How high above the top of the loop must it be released in order that the ball just makes it around the loop?

    Repeat problem (a) for a disk. Find the ratio of the heights h for the two cases.

    I used conservation of energy and set mgh = KE(rot) + KE(trans) + mg2R. Using V=sqrt(Rg), I found the minimum speed the ball must have at the top of the loop and found it to be 5.512 m/s. For the right side of the equation, I used KE(rot) = 1/2Iw(omega)^2. For a spherical ball, I =2/5MR^2.
    Now I have
    mgh = 1/2mv^2 +1/2(2/5)MR^2(w^2) + 2mgR
    M's cancel
    w = v/R
    So I was left with
    h = (1/2(v^2)+1/5(R^2)(v/R)^2 + 2gR)
    h = 8.37m
    h-2R = H = 2.17m

    That is the correct answer according the homework but then it asks for a disk with the same radius and here is where I don't get the right answer.

    mgh = 1/2mv^2 +1/2Iw^2 + mg2R
    mgh = 1/2mv^2 +1/2*1/2mR^2w^2 + mg2R
    m's cancel

    h = 1/2v^2 +1/4R^2(v/R)^2 +2gR

    Help anyone?
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted

Similar Discussions: Loop the Loop