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Homework Help: Rock on Strings driving me crazy!

  1. May 28, 2010 #1
    Physics 101 (forces), Rock on Strings: driving me crazy!

    okay, this problem is DRIVING ME CRAZY!! please help me out here, it's a TYCHO homework:

    1. The problem statement, all variables and given/known data

    A 540 gram rock is held by two strings of 45 cm with ends 57 cm apart (distance d in the picture), any string will break under a tension of 32 N. The rock is whirled in a circle between them. Neglecting gravity: what is the maximum speed the rock can have before one of the string breaks?

    the value I got is V = 5.164 m/s but it's WRONG and I cannot find any mistake.

    the picture of this problem would be the following one (picture d):
    [PLAIN]https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/summer/homework/Ch-06-Forces/rock_on_strings/13.gif [Broken]

    3. The attempt at a solution

    1* First I calculated the radius of the circle (R in the picture): r = 0.3482 m

    2* Then I found the angle between the radius and one of the strings: angle = 39.30 degrees.

    3* I calculated what the max. value of the acceleration in the opposite direction of the tension would be using the formula F = m x a, this is:
    32N = 0.54kg x Ay
    32N / 0.54kg = Ay
    59.26m/s^2 = Ay

    (Ay would be acceleration in the Y direction, the direction of the tension)

    4* since I got the Y component of the acceleration, I used angle in point 2* to calculate the value of the total acceleration:

    cos(39.30) = Ay / A
    A = Ay / cos(39.30)
    A = 59.26 / cos(39.30)
    A = 76.58m/s^2

    5* Okay so now I have the maximum acceleration that one string could handle. To calculate the velocity I simply use the formula A = V^2 / r:
    V = sqr(A x r)
    V = sqr(76.58 x 0.3482)

    V = 5.164 m/s


    please let me know if you see any mistake in my calculations or if I didnt set the problem out properly!!
    many thanks in advance!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 28, 2010 #2
    A picture would be very helpful and would increase the number of responses you get. Right now I am unable to picture what this setup looks like.
  4. May 28, 2010 #3
    thanks! you're right, I dont have much experience in asking for help for my homework, but there it is, the picture of the problem.
  5. May 28, 2010 #4
    First, you shouldn't be calculating the acceleration in the string since that doesn't make much physical sense. Also, the tension isn't only in the y-direction.

    To start you off, I recommend drawing a free body diagram for the mass. I will give you a hint, there should only be two forces acting on the mass (since we neglect gravity). Since the ball is not moving in the y-direction, the y-component of the net force will be 0. And since the ball is moving in a circular motion, then the x-component of the net force will be equal to the centripetal force keeping it in the circular motion.

    Solving for those two equations should lead you to the correct answer.
  6. May 28, 2010 #5
    well, I understand what you are saying and that's exactly what I did.... I draw an Inclined plane in which the direction of Y is the direction of the upper string, and the origin would be right in the mass. The acceleration I was talking about was the centripetal acceleration, which, if you look at the picture, would be exactly horizontal, but looking at the inclined plane it would point to the fourth quadrant.
    That being said, the only component of the acceleration (centripetal acceleration) that I'm interested in would be the Y component because it's the only one occurring in the direction of the string.
  7. May 29, 2010 #6


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    No, you took the string to be the y-component of some acceleration. But this doesn't make sense; there's no x-component. Looking at it another way, if I plug 90 degrees into your equation (corresponding to the strings being collinear), I get a rock acceleration of infinity, which doesn't make any sense. (Hint: the correct answer should be zero for this configuration.)

    nickjer's suggestion, that you calculate the rock's acceleration and divide that between the strings (using the correct trigonometric relationship) is the most intuitive way to solve the problem and will lead you to the right answer. Alternatively, you can divide the string tension into orthogonal components, one of which accelerates (half) the rock and the other of which pulls against the other string. But right now you're essentially dividing by a cosine when you should be multiplying by one.
  8. May 29, 2010 #7
    alright thank you nickjer and mapes!! correct answer was 5.651 m/s (I wasn't THAT wrong ^_^), I only needed another way to see the problem! thanks guys
  9. May 29, 2010 #8
    "There should only be two forces acting on the mass."
    "Alternatively, you can divide the string tension into orthogonal components."
    "But right now you're essentially dividing by a cosine when you should be multiplying by one."

    those were the key points :)
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