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Radial Velocity

  1. Nov 10, 2003 #1
    I had a problem on my physics test today that said something like. You have a string with radius of 25 or 50 cm (cant remember which) and you have a mass hung on the end of the string. And the intial velocity is zero when the string is horizontal. The question was what is the velocity at the bottom of its circular path. How can you solve this? I was stumped, it was a multiple choice question and the answers were approximate numbers ranging from 2 m/s to 6 m/s. I just do not see how you could solve this problem without knowing the mass, and getting an approximate number. Also it did not say that the strings mass was neglible. Any clues?
  2. jcsd
  3. Nov 10, 2003 #2


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    Well, this question is bordering on "homework help," but, since it's about a concept, I'll let it remain here.

    The easiest way to solve this problem is via the conservation of energy. The ball loses gravitational potential energy as it descends, and gains the same amount of kinetic energy.

    Gravitational potential energy lost is m g y, where m is the mass of the object and y is the (vertical) distance it fell.

    Kinetic energy is expressed as 1/2 m v2.

    Equating the loss of gravitational potential energy and the gain of kinetic energy look like this:

    1/2 m v2 = m g y

    And you can simply solve for the velocity:

    v = sqrt(2 g y)

    Does this make sense?

    - Warren
  4. Nov 10, 2003 #3
    Yeah I was wondering if I should have put it here. And yes that makes complete sense, thanks. The mass cancels out, geeze I even had that v=sqrt(2gy) forumla written down. I guess I just gotta hope I guessed right
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