Conservation of mechanical energy for skier on sphere

[NathanLeduc1]

Homework Statement

A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. At what angle θ will the skier leave the sphere?

Homework Equations

KE= 0.5mv^2
PE = mgh
Fc = (mv^2)/r

The Attempt at a Solution

I am really quite confused and don't even know how to begin. I know that the skier will fall off when the normal force is 0 but I'm not sure how to even get to that point.
I thought that PE=KE so 0.5mv^2 = mgh which simplifies to 0.5v^2 = gh

 

[Doc Al]

I know that the skier will fall off when the normal force is 0 but I'm not sure how to even get to that point.

That's the key point. To make use of it, set up a force equation with Newton's 2nd law.

I thought that PE=KE so 0.5mv^2 = mgh which simplifies to 0.5v^2 = gh

Good. You'll need that too. (Express h in terms of θ.)

 

[NathanLeduc1]

So rewriting the energy equations gives me
v^2 / 2 = grsin(θ).

When I write the force equations, I get
N-mg = ma
N = mg + mg
N = m(a+g)
m = N/(a+g)

If I plug that into the centripetal force equation, I get
F = (mv^2)/r
F = (Nv^2)/((r)(a+g))

Is this right? If so, where do I go from here? Sorry to ask such dumb questions, I'm just very confused on this problem.

 

[Doc Al]

So rewriting the energy equations gives me
v^2 / 2 = grsin(θ).

What you want is Δh, the drop from the original position at the top. Δh ≠ r sinθ.

When I write the force equations, I get
N-mg = ma
N = mg + mg
N = m(a+g)
m = N/(a+g)

Careful! Forces are vectors.

Hint: Consider force components perpendicular to the surface at any point.

 

[Nickie]

, but I don't know how to incorporate the radius of the sphere into this equation.

I would like to clarify that the conservation of mechanical energy principle states that the total mechanical energy (potential energy + kinetic energy) of a system remains constant as long as there are no external forces acting on it. In this case, the skier-sphere system is isolated, meaning there are no external forces acting on it, so the total mechanical energy will remain constant.

To solve for the angle θ at which the skier will leave the sphere, we can use the conservation of mechanical energy principle and equate the initial mechanical energy (when the skier is at the top of the sphere) to the final mechanical energy (when the skier is at the point of leaving the sphere).

At the top of the sphere, the skier's potential energy is mgh, where h is the height of the skier from the ground. Since the skier is at rest, the initial kinetic energy is 0.

At the point of leaving the sphere, the skier's potential energy is 0, since the skier is at ground level. The final kinetic energy is given by 0.5mv^2, where v is the speed of the skier at that point.

Since the total mechanical energy remains constant, we can equate the initial and final mechanical energies to get:

mgh = 0.5mv^2

We can then rearrange this equation to solve for the speed v:

v = √(2gh)

Now, we can use this value of v in the equation for centripetal force (Fc = (mv^2)/r) to solve for the radius r of the sphere:

Fc = (mv^2)/r

Substituting the value of v we just found, we get:

Fc = (m√(2gh)^2)/r

Simplifying, we get:

Fc = (2mgh)/r

Since the skier will leave the sphere when the normal force (Fc) is 0, we can set this equation equal to 0 and solve for the angle θ:

0 = (2mgh)/r

Solving for θ, we get:

θ = sin^-1 (r/2h)

Therefore, the skier will leave the sphere at an angle of sin^-1 (r/2h).