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Mechanical Energy:Potential, kinect, etc.

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A sphere of mass 2kg is release from point A off the edge of a hole that is represented of as a semi-circle of R=0,2 m.
    consider g=10m/s²
    What is the force that the ground apply on the sphere when it passes trough point B(bottom of the hole?

    2. Relevant equations
    Ep=m*g*h
    Ec=m*v²/2
    Q=m*v
    F=m*a
    3. The attempt at a solution
    I tried finding the Potential energy of the sphere at point A that is
    Ep=2kg*0,2m*10m/s² = 4J
    and the weight of the sphere that is 2*10= 20 N
    I don't know what to do anymore the velocity at point B i considered 2m/s because of
    Ep=Ec=m*v²/2 then V=2m/s
    Help?
    Image not from actual problem point A would be the left point B where it is on the pic
    35c+Stable+Ideal+copy.jpg
     
    Last edited: Dec 5, 2013
  2. jcsd
  3. Dec 5, 2013 #2

    ShayanJ

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    Gold Member

    You can simply find the speed of the sphere when it reaches point B using conservation of energy.Call it [itex] v_B[/itex]. When the sphere is moving in the semi-circle,it needs a centripetal force which at point B is equal to [itex] \frac{mv_B^2}{R} [/itex]. But this centripetal force can be provided only by the circular track and nothing else so your answer is [itex] \frac{mv_B^2}{R} [/itex].
     
  4. Dec 5, 2013 #3
    AH thanks the answer is 60N i guess you have to sum that with the weight that is 20N
    40 +20 =60 N
    but i dont get where mv2/r comes .. i dont have that on my theory
    they just gave mv²/2 and m*g*h
     
  5. Dec 5, 2013 #4

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    You need to understand the centripetal force concepts before solving this problem. You should know that the centripetal force mv^2/r is the net force acting inward toward the center of the circle. So the centripetal force is 40 N inward and is comprised of the net force of the ground normal force acting up
    and the weight acting down, the net force
     
  6. Dec 5, 2013 #5
    Thank you, now i see why couldnt soulve any problems envolving circles xD
     
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