What is the relationship between h and R in this circular motion problem?

AI Thread Summary
The discussion centers on determining the height h at which a skier leaves a frictionless hemispherical dome of radius R. The skier starts from rest, and the normal force becomes zero at height h. Participants suggest using conservation of energy and centripetal acceleration to derive the relationship between h and R, with hints to draw free body diagrams and consider forces acting on the skier. The correct answer is identified as h = 2R/3, but confusion arises regarding the application of centripetal force and the components of gravitational force. Ultimately, understanding the forces at play and visualizing the problem through diagrams is emphasized as crucial for solving it.
Almond
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A skier starts from rest at the top of a frictionless slope of ice in the shape of a hemispherical dome with radius R and slides down the slope. At a certain hight h, teh normal force becomes zero and the skier leaves the surface of teh ice. What is h in term of R?

I know the answer is 2R/3, but couldn't figure out how to get there.

Thanks.
 
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Show what you've done so far and where you are stuck.

Hint: Consider the forces acting on the skier at any point and apply Newton's 2nd law.
 
Welcome to PF!

Hi Almond ! Welcome to PF! :smile:
Almond said:
… At a certain hight h, teh normal force becomes zero

Hint: centripetal acceleration :wink:
 
Well, I tried to use conservation of energy,
mgR=mgh+1/2mv2
Then I used centripetal acceleration, a=V2 /R
gR=gh+1/2aR,

And I'm stuck, not sure how to get a.
 
ok, now draw a diagram (for yourself) and call the angle θ …

what is the normal force when the skier is at angle θ from the vertical? :smile:
 
another hint:

Draw a free body diagram of the skier as he leaves the slope. What is the normal force at that point? what are the sum of the forces in the direction of the normal force? The skier "drops" from R to h, is there an equation of motion that relates velocity as a function of position? The geometry of the skier as he leaves the slope in terms of R and h will be helpful.
 
I have managed to get the answer somehow by using these two equations:

h=Rsin\theta and
a=gsin\theta

But I don't understand the second equation, shouldn't it be g=asin\theta, since mg points downwards and is the component of centripetal force, I suppose? I am totally confused with the free body diagram.:frown:

Thanks for all the help so far!:smile:
 
Almond said:
But I don't understand the second equation, shouldn't it be g=asin\theta, since mg points downwards and is the component of centripetal force, I suppose? I am totally confused with the free body diagram.
If mg points down, what is its radial component?

Draw a diagram at the moment that the skier just begins to lose contact with the surface. What forces act? What are their components? What net force acts toward the center of the sphere?
 
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An approach that helped me solve this problem was drawing a free body diagram for the skier, from the point of view of his accelerated system and look at all the forces he has acting upon him. From there, just look at the specific instant where N=0 and find the correlating angle Θ from the vertical.
 
  • #10
I agree with RoyalCat. Whenever I come across a problem involving centripetal acceleration, I always pretend there's a centrifugal force acting on the accelerated body. It makes so much more sense for me.
 
  • #11
and I could never work my way with centrifugal force.. always wondered, "where is this coming from!?"
 
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