Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

GAMSAT Sample Paper Unit 2

Options
  • 23-08-2011 8:26pm
    #1
    spotsanddots
    Registered Users Posts: 132 ✭✭


    Any clues on how to go about Q8.
    A spherical ball radius 1cm mass 50kg rolls down a curved surface radius 8cm.the gravitational acceleration of a free falling object is 10ms^-2.
    What is the decrease in gravitational potential energy of the ball as it rolls?

    I thought it would just be the gase of Ep=mgh but its not..any ideas?


Comments

  • Options
    #2
    Bonnieaurstomp
    Registered Users Posts: 24 ✭Bonnieaurstomp


    spotsanddots wrote: »
    I thought it would just be the gase of Ep=mgh but its not..any ideas?

    It is, but you need to measure the vertical drop of the ball measuring from the centre point of the ball. And make sure you convert the distance to meters.


  • Options
    #3
    spotsanddots
    Registered Users Posts: 132 ✭✭spotsanddots


    the vertical drop would be 8cm (0.08m) would that just be 1cm less if the centre pt is in the centre of the ball?
    I thought it was 50kg x 10ms^-2 x 0.08m= 32
    The actual answer is 3.5x10-2 and since its multiple choice you are told to choose the nearest answer am i supposed to divide by 1000 for some reason?


  • Options
    #4
    Annaroberts22
    Closed Accounts Posts: 29 ✭Annaroberts22


    spotsanddots wrote: »
    the vertical drop would be 8cm (0.08m) would that just be 1cm less if the centre pt is in the centre of the ball?
    I thought it was 50kg x 10ms^-2 x 0.08m= 32
    The actual answer is 3.5x10-2 and since its multiple choice you are told to choose the nearest answer am i supposed to divide by 1000 for some reason?

    Hi :-)

    The units have to all be the right ones for the equation. Mass in Kg and height in m. So (0.05 x 10 x 0.1) is the potential energy at point R and (0.05 x 10 x 0.03) at point T. What matters is where the centre of mass is so at R the centre of mass is 10 cm off the ground and at T it is 3 cm off the ground and the difference (0.05 x 10 x 0.07) is the potential energy change.


  • Options
    #5
    spotsanddots
    Registered Users Posts: 132 ✭✭spotsanddots


    thanks so much im such an idiot would have had the right answer had i looked at the units correctly:mad:


Advertisement