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Does circular swinging increase the distance of a throw?

  1. Jun 19, 2013 #1
    It is like the throws in hammer throw, when an object is swung circularly, the it does contains an rotational energy right? when it is release, how the energy affect the objects? Does the energy change to velocity tangent to the circular path it swung?
     
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  3. Jun 19, 2013 #2

    Bandersnatch

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    The point in swinging is to increase the initial velocity of the object at the moment of release. That velocity is equal to the instantenous tangential velocity in circular motion. Swinging around allows the thrower more time to accelerate the object.
    Initial velocity, of course, translates into increased distance travelled, as per the equations of motion.

    If you'd rather think of it in terms of energy, then swinging around means that there is virtually an unlimited path(a closed circle) on which a relatively low force supplied by the thrower can perform work on the system(thrower+object) to increase its rotational energy.
    At the moment of release, the rotational energy is transfered to the two components of the system(thrower+object) as kinetic energy. Higher kinetic energy means higher initial velocity, which, again, means farther throw.
    Compare with throwing just by using one's arm's length - it would require much higher force to impart the object with the same amount of energy, as the distance over which the work can be performed is very limited.
     
  4. Jun 19, 2013 #3
    Understood, thanks
     
  5. Jun 19, 2013 #4

    rcgldr

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    The purpose of this process is to increase the objects linear kinetic energy (speed^2). Energy = force x distance, and the distance in this case is a unlimited path.

    The rotational energy won't increase the distance of a throw directly, other than if there's backspin, you could get some amount of aerodynamic lift (magnus effect).
     
  6. Jun 19, 2013 #5

    Bandersnatch

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    Isn't that what I said in the next sentence?
     
  7. Jun 19, 2013 #6

    rcgldr

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    It wasn't clear if you meant the rotational energy with respect to the center of the circular path of the object (KE = 1/2 m R^2 ω^2, where R = radius of circular path), or the rotational energy of the object itself (KE = 1/5 m r^2 ω^2, where r = radius of object).
     
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