Boy Seated On Mound of Ice, Starts Sliding

[yrsnkd]

1. A boy is seated on the top of a hemispherical mound of ice with radius R. He is given a very small push and starts sliding down the ice. At what height does he leave the ice if the ice is frictionless?



2. Conservation of Mechanical Energy: U₁ + K₁ = U₂ + K₂; also gravitational energy mgy and kinetic energy (1/2)mv²



3. The normal force vanishes when the boy leaves the ice; up until that point, however, the normal force pushes him with a force that changes in magnitude and direction with his sliding off. Since work is the integral of a force over a distance, this is where the boy must get the kinetic energy from, besides the gravitational force. Past that, though, I have no idea what to do.

 

[Fewmet]

1. A boy is seated on the top of a hemispherical mound of ice with radius R. He is given a very small push and starts sliding down the ice. At what height does he leave the ice if the ice is frictionless?

3. The normal force vanishes when the boy leaves the ice; up until that point, however, the normal force pushes him with a force that changes in magnitude and direction with his sliding off. Since work is the integral of a force over a distance, this is where the boy must get the kinetic energy from, besides the gravitational force. Past that, though, I have no idea what to do.

That's a neat problem, and the approach is not immediately obvious to me. Have you tried a force diagram? I think you can treat him at every point as though he were on an incline tangent to the surface. Do you know the general form for the normal force and net for for a body sliding down a frictionless incline?

 

[PhanthomJay]

The normal force vanishes when the boy leaves the ice; up until that point, however, the normal force pushes him with a force that changes in magnitude and direction with his sliding off. Since work is the integral of a force over a distance, this is where the boy must get the kinetic energy from, besides the gravitational force. Past that, though, I have no idea what to do.

The normal force does no work, so it can't change the KE. You might want to try applying your conservation of energy equation between the known point and unknown point where N = 0, to give one equation, then try using the centripetal force equation at that point for a 2nd equation.

 

[WatermelonPig]

mgcos(theta) = m[(v^2)/R]
mgcost(theta) = m[(v^2)/R]
gcos(theta) = (v^2)/R
Where theta is the angle for the tangent line. I think this should be correct.

 

[AlexChandler]

If \theta is the angle between the gravitational force vector on the boy and the position vector from the center of the sphere to the boy, then the height of the boy is always h \cos \theta where h is the radius of the sphere.
At the time the boy leaves the sphere, the centripetal force is mg \cos \theta
and this is equal to centripetal acceleration.
Also consider conservation of mechanical energy to find that
mgh = 1/2 m v^2 + mgh \cos \theta
These relations should help