At What Point Does the Boy Lose Contact with the Ice?

In summary, the boy slides off the ice at an initial speed of 3m/sec and loses contact with the surface at an angle of 12.26°. His speed now is 0m/sec and he needs a centripetal force of 0.652m/s2 to just maintain contact with the ice.
  • #1
NilliNilli
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Homework Statement



A boy is initially seated on the top of a hemispherical ice mound of radius r=13.8m. He begins to slide down the ice, with an initial speed vi=3m/sec. Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?


Homework Equations





The Attempt at a Solution



My instructor gave the hint that it has to do with how force is relevant to centripetal force.
I first attempted finding the boy's direction and speed right as he started sliding off the ice by doing vector addition with the initial velocity (which I interpret as his speed in the x direction) and the centripetal acceleration ( interpreted as his speed in the y direction).
I found the centripetal acceleration to be v2/r = (3)2/13.8 = 0.652m/s2. Then I found the initial angle θ=12.26°.

I kind of abandoned that since I didn't think I was approaching it the right way at all. Going off of my instructor's hint, I figured the boy has to start off with a centripetal force, since his acceleration is centripetal to start with, then since he's not attached to the sphere his acceleration become free-fall. I'm not sure how I should go about calculating at what point his acceleration stops begin centripetal. I'm thinking I have to determine when that happens then use s=rθ to find the arc length, since I'm not sure how else to determine distance on the sphere.
 
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  • #2
I would use energy concepts here. But in terms of forces (Newton's second law) how do you know the boy loses contact with the ice? Try expressing that critical condition first
 
  • #3
That's what I need help with. I need to express that in numbers and equations and I don't know how to do that for this problem.
 
  • #4
Well, typically when an object loses contact with any surface, the Normal force becomes less than or equal to 0. So draw a force diagram and label all the forces on the boy first
 
  • #5
That was the first thing I did, actually.
 
  • #6
So what equation did you get using Newton's second law and the forces you found?
 
  • #7
F=mg+n
I thought if the force was centripetal it could be
F=mg+n=mv2/r
I understand what you mean that when n=0, that will be when the boy lost contact with the ice, but I'm still not sure how to connect that to an equation where I can find a time or height that it happens.
 
  • #8
Are you sure you have your signs correct? Where is the Normal force pointing and where is the force of gravity pointing?
 
  • #9
I have normal pointing up and gravity pointing down, assuming this is at the top of the sphere. Assuming this is right where the boy's about to lose contact, gravity would still be down and normal force would be pointing perpendicular to the sphere. The normal force's direction will consistently change until it's 0.
 
  • #10
Suppose he loses contact when he has traveled through an arc of angle theta. Through what height has he descended? What is his speed now? What centripetal force is needed there to just maintain contact? What forces are available to provide it?
 

1. What is the significance of the boy sliding off of the ice mound?

The boy sliding off of the ice mound is a common physics demonstration used to explain the concepts of friction and momentum. It shows how friction can slow down the momentum of an object, in this case the boy, as he slides down the icy surface.

2. Why is the boy sliding off of the ice mound instead of staying in place?

The boy is sliding off of the ice mound due to the force of gravity acting on him. As he starts to slide, the force of friction between his body and the icy surface slows him down, but it is not enough to completely stop his movement.

3. What other factors could affect the boy's sliding on the ice mound?

The speed at which the boy slides, the angle of the ice mound, and the texture of the ice surface can all affect the boy's sliding on the ice mound. These factors can impact the amount of friction and momentum acting on the boy.

4. Is the boy's sliding on the ice mound an example of Newton's Laws of Motion?

Yes, the boy's sliding on the ice mound can be explained by Newton's Laws of Motion. The first law states that an object will remain in motion unless acted upon by an external force, in this case the force of friction. The second law explains how the amount of force and acceleration are related, and the third law states that for every action there is an equal and opposite reaction.

5. How does the boy's sliding on the ice mound relate to real-world applications?

The boy's sliding on the ice mound relates to real-world applications in sports, such as ice skating and skiing, where understanding friction and momentum is important for performance. It also has practical applications in engineering and design, such as creating surfaces with specific friction levels for safety purposes.

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