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Maximum kinetic energy from a circular movement.( 1997 /1996 Leaving Cert)

  • 22-11-2005 1:59pm
    #1
    ublinia2
    Closed Accounts Posts: 114 ✭✭


    I think this question was on either the 97, 96 or earlier Physics Leaving Cert. It is a mechanics question.

    The question shows a 'pendulum' movement with an angle of displacement from the centre /equilibrium point of the pendulum as 35 degrees. ( As a result you can only use circular motion formula and not simple harmonic motion formula)

    The length of the string is 0.85m from the point of suspension to the bob.The question asks the student to calculate the maximum kinetic energy created from the position the bob is in.

    I tried to do the question using vectors but want to check if the method is correct. Can somebody have a look at it .Hopefully I will post up the question later.

    Thanks


Comments

  • #2
    dudara
    Registered Users, Registered Users 2 Posts: 33,518 ✭✭✭✭dudara


    This is a conservative system, so the potential energy gained by the bob as it is raised up, will be converted into KE. So what they're looking for you to do here, is calculate the height that the bob is raised through, use this to calculate PE and then explain that in a conservative system, this will be converted to KE.

    Have a bash in calculating it. You should get h = 0.15 m and the rest is straightforward. ( I hope)


  • #3
    ublinia2
    Closed Accounts Posts: 114 ✭✭ublinia2


    Thats what I thought myself , but how exactly do you calulate the ball is at -
    and the change in height as the ball falls ??

    I dont think its that easy ?


  • #4
    Delphi91
    Registered Users, Registered Users 2 Posts: 1,501 ✭✭✭Delphi91


    ublinia2 wrote:
    Thats what I thought myself , but how exactly do you calulate the ball is at -
    and the change in height as the ball falls ??

    I dont think its that easy ?

    Yes, Dudara is right - it is a conservation of energy problem. The P.E. which the pendulum has at the start is converted totally to K.E. when the pendulum is released.

    So, mgh = 1/2(mv^2)

    or gh = 1/2(v^2)

    If you re-arrange, you get

    v = sqr root(2gh)

    where h = 0.85 - (0.85 * cos 35)

    Have a look here.

    Hope this helps.

    Mike


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