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Changing radius and centripetal force

  1. Oct 12, 2009 #1
    I have my own concept question.

    Ignoring air friction and mass of string, and only considering the motion of the bottom right quarter of a circle, where only gravity acts on the mass.

    Lets say I have a mass attached to a rope with length R. I hold the string in my left hand and the mass in the other hand. I let go the mass so it will fall in a circular motion and be vertical at the bottom my left hand.

    Now I have a yo-yo with the same mass and with initial R as well. I do the same thing with the yo-yo as I did with the first mass. The yoyo will be changing length as it falls.

    Will the yo-yo and the first mass take the same time to reach the bottom? I'm guessing yes because they have the same tension and only force acting on it is gravity. Meaning, they will have the same velocities.
     
  2. jcsd
  3. Oct 12, 2009 #2

    rl.bhat

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    Initially both will have the same potential energy.
    In the case of falling mass, at any point the total energy = PE + KE
    In the case of yo-yo, at any point the total energy = PE + KE + rotational KE.
    Hence at any point linear velocity of mass is greater than the velocity of yo-yo.
     
  4. Oct 12, 2009 #3
    Interesting, never thought of using energies. But if thats the case, how do we know the yo-yo will be slower for sure. Since the yo-yo is lower at the bottom, it has less PE and so higher KE than the mass? If we ignore the rotational KE of the yo-yo, but still include the increasing length, does this mean for sure the yo-yo will reach the bottom first then?

    But if use the equation for centripetal force at the bottom

    Using same T-MG for net force.

    Fnet = mv2/r = 4pi2r/time.

    time=4pi2r/Fnet

    Meaning greater R will take longer, thus yo yo should take longer? Meaning i'm wrong about the top assumption?
     
    Last edited: Oct 12, 2009
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