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Spring Motion (1st Semester Physics)

  1. Nov 8, 2004 #1
    Hey guys,

    Got a problem here that is getting quite annoying...

    A spring is 0.6 m long. One end is permanently attached to a pivot on a horizontal table top. The force that it exerts on a body attached to the other end is (4.2 N/m)x, where x is the distance that the spring is stretched beyond its normal length. Suppose a mass of 400 g is attached to the other end of the spring and the whol system is set in circular motion. (The entire motion takes place on the horizontal table top.) How far will the spring be stretched if the mass rotates with an angular velocity of 8 rad/s?

    All right, first off a = m(w^2)r in circular motion, where m is the mass, w (omega) is the angular velocity and r is the radius. However, in this equation the radius is .6 + x, where x is the change in distance. Since this is the only force involved in the problem, it must be going toward the center of the circle (because centripetal acceleration is only the result of other accelerations). This means that 4.2x = m(w^2)r. However, this leaves me with,

    4.2x = .4(8^2)(.6+x)
    (4.2x)/(.6+x) = 25.6
    (4.2x)/(.6+x) - 25.6 = 0

    Problem is, this does not yield a correct answer...

    Then I tried again using mv^2/r = a and wR=v... this yielded 1.2 meters, the answer should be .87 meters. I know there's an error somewhere, I just can't find it. Can anybody lend me a hand?

    Thanks in advance,
  2. jcsd
  3. Nov 8, 2004 #2

    Still having the same problem, but I realized why my acceleration from v and my acceleration from w weren't the same, simply because I forgot to leave r as .6 + x. When I put this in it gives me the same answer as the w formula, though the answer is still wrong. Thoughts?

    Thanks in advance,
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