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

In summary, at a certain height h, the normal force becomes zero and the skier leaves the surface of the ice. The normal force is mg points down and its radial component is unknown.
  • #1
Almond
3
0
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.
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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:
 
  • #4
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.
 
  • #5
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:
 
  • #6
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.
 
  • #7
I have managed to get the answer somehow by using these two equations:

h=Rsin[tex]\theta[/tex] and
a=gsin[tex]\theta[/tex]

But I don't understand the second equation, shouldn't it be g=asin[tex]\theta[/tex], 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:
 
  • #8
Almond said:
But I don't understand the second equation, shouldn't it be g=asin[tex]\theta[/tex], 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?
 
Last edited:
  • #9
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!?"
 

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

What is a circular motion problem?

A circular motion problem is a type of physics problem that involves an object moving in a circular path. It often requires the use of equations and concepts related to circular motion, such as centripetal force and angular velocity.

How do you solve a circular motion problem?

To solve a circular motion problem, you need to first identify the known and unknown variables. Then, use equations related to circular motion, such as the centripetal force equation or the angular velocity equation, to solve for the unknown variable.

What are some real-world examples of circular motion problems?

Examples of circular motion problems in real life include the motion of a planet around the sun, a car going around a roundabout, or a ball spinning on a string.

What is the difference between uniform circular motion and non-uniform circular motion?

In uniform circular motion, the speed of the object remains constant while the direction changes. In non-uniform circular motion, the speed and direction of the object are constantly changing.

How does centripetal force affect circular motion?

Centripetal force is the force that keeps an object moving in a circular path. Without centripetal force, the object would move in a straight line. The magnitude of the centripetal force is directly proportional to the mass and speed of the object and inversely proportional to the radius of the circular path.

Similar threads

Back
Top