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Need help with cons. of energy problem

  1. Nov 15, 2004 #1
    The problem is a boy is seated at the top of a hemispherical mound of ice so theres no friction. The radius of the mound is R. He is given a very small push so at the top there is basically only potential energy with a height of R. After he starts sliding down i need to show that the point where he leaves the ice is at a height of 2R/3. I know this point would be where the normal force is equal to zero.
    Since the normal force is zero i set the centrepital force equal to the weight at that given moment but im not sure if thats right. Any help is greatly appreciated.
     
  2. jcsd
  3. Nov 15, 2004 #2

    Tide

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    The boy will leave the surface when the centripetal acceleration is equal to the component of acceleration due to gravity that is normal to the surface.
     
  4. Nov 15, 2004 #3
    thanks, i understand that but i dont know how to find the component of gravity that is normal to the surface with no angles given.
     
  5. Nov 15, 2004 #4
    well, when the boy is on an incline, is the normal force equal to the full weight of the boy? or to a component thereof, perhaps? how about this: try drawing a triangle from the center of the (semi)circle to the boy's position, down perpendicular to the ground. can you express the component of the boy's weight in the direction of N in terms of an angle of that triangle? can you, in turn, express that angle in terms of other quantities you know/are working with (such as height or radius or whatever)?

    anyway, draw a careful diagram with forces on it, will help lots
     
  6. Nov 15, 2004 #5
    i tried drawing a triangle but i dont think u can get the angle because if u draw a line straight down from the position of the boy which would be equal to his weight you cant be sure that the triangle will have a right angle in it anywhere. With no right angle i dont think u can find the component of the weight in terms of the weight.
     
  7. Nov 15, 2004 #6
    and if u just draw the line down to the ground forming a right angle, how would u know that is equal to the weight
     
  8. Nov 15, 2004 #7
    well, if you draw a straight line down to the ground from the boy's position, it's bound to be perpendicular to the ground, isn't it? that's what I mean by straight - right in the direction of the boy's weight. and, incidentally, that side of the triangle will also be equal to the height of the boy at that point in time, which is a variable in your calculation
     
  9. Nov 15, 2004 #8
    i appreciate the help but im still confused. that side of the triangle is what im trying to find yes, the height at that point in time but i dont see what i can set it equal to. if i throw in an angle, wont it just give me another unknown variable.
     
  10. Nov 15, 2004 #9
    if you've a scanner, show me the diagram you've got. basically, you'll need the sine of that angle, which you can express in terms of R and y (the height). so no new variables.
     
  11. Nov 15, 2004 #10
    i dont have a scanner on this comp but give me one second and ill draw my diagram real quick and attach it
     
  12. Nov 16, 2004 #11
    trying to make the picture smaller so i can attach it
     
  13. Nov 16, 2004 #12
    alright, heres my diagram
     

    Attached Files:

  14. Nov 16, 2004 #13
    allright, diagram looks fine. now if you draw the component of the boy's weight in the direction of the normal force, it'll be exactly on the line connecting the boy and the center of the semicircle. make another small triangle with the full weight as the hypothenuse (sp?) and the components as the (what do you call the other 2 sides in English? only know it in Latvian). this triangle will be similar to the big one and the angle pheta will appear in it too. one more thing to note is that sin(pheta)=(h/R). Your goal now is to find where the component of the boy's weight in the normal direction becomes equal to the centripetal force - express the velocity in the centripetal force in terms of height (from conservation of momentum) and set the two things (component and centripetal force) equal.

    work with it for a while to see if you can figure out what i mean
     
  15. Nov 16, 2004 #14
    Perfect explanation, i really appreciate your help
     
  16. Nov 16, 2004 #15
    if ur still here, although it makes sense, when i try to solve it i cant get h = 2R/3.
    i did END = START
    then the final potential energy + Fc = initial potential energy
    so my eq. looks like mgh + mgR/h = mgR

    i dont know if thats right but would appreciate your help one more time
     
  17. Nov 16, 2004 #16
    well, all you need to know that the change in potential energy is equal to the change in kinetic energy. that is, mgy - where y is the _change_ in height - is equal to mv^2/2. From that you can extract v in terms of the distance fallen.
     
  18. Nov 16, 2004 #17
    [tex]\tau=mgRsin{\Theta}[/tex]
    [tex]I\frac{d\omega}{dt}=mgR\sin{\Theta}[/tex]
    [tex]\omega\frac{d\omega}{d\Theta}=\frac{g\sin{\Theta}}{R}[/tex]
    After integration you should be able to get the general formula for the angular speed of this case,
    [tex]\frac{\omega^2}{2}=\frac{g}{R}(1-\cos{\Theta})[/tex]
    Sub in [tex]\cos{\Theta}=\frac{h'}{R}[/tex] and you should get
    [tex]\omega^2=\frac{2g}{R^2}(R-h')[/tex]
    By equaling [tex]mR\omega^2=mg\cos{\Theta}[/tex] and sub in all the necessary things, you will eventually get the answer.
     
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