# Homework Help: Finding the angle of a falling person on a sphere

1. Sep 3, 2007

### Enderless

1. The problem statement, all variables and given/known data

A sphere is on the ground, with the sphere having a radius R. A person (skier) is standing on top of the sphere. neglecting friction, find the angle at which the person will slide off the sphere if he moves from rest to a constant velocity of v.

http://img63.imageshack.us/img63/930/untitledai7.jpg [Broken]

2. Relevant equations

F = ma
F = mv^2/r (centripetal force)
U + K = U + K (conservation of energy)

3. The attempt at a solution

1/2mv^2 + mgh = 1/2mv^2 + mgh
0J + m(9.8m/s^2)(2R) = 1/2mv^2 + m(9.8m/s^2)(R + Rcos (angle))

So that simplifies to v^2/2 = 9.8 m/s(R) - 9.8m/s(Rcos (angle))

Drawing a free body diagram, I know that the centripetal force is equal to the Normal foce*cos(angle):

mv^2/r = mgcos(angle)
v^2/r = 9.8m/s (cos (angle)
v^2 = 9.8 m/s (Rcos(angle))

Substitute that into v^2/2 = 9.8 m/s(R) - 9.8m/s(Rcos (angle)) and get:

9.8m/s Rcos (angle)) = 19.6 m/s(R) - 19.6m/s (Rcos (angle))

That reduces to:

cos (angle) = 2/3
so the angle must be 48.2 degrees

My teacher said it was wrong, can anyone assist me?

Last edited by a moderator: May 3, 2017
2. Sep 3, 2007

### HallsofIvy

I a person is sking down a sphere, there is no "centripetal force"- there is no force on the person toward the center of the sphere. The only force is that of gravity, straight down.

3. Sep 3, 2007

### Enderless

A centripetal force is included in the problem. I didn't type the whole problem out.

4. Sep 3, 2007

### learningphysics

The centripetal force is the force towards the center due to gravity and the normal... if the skier is moving along the sphere:

the net force towards the center = mv^2/r

component of gravity towards the center - normal force = mv^2/r

When the component of gravity towards the center becomes less than mv^2/r it is impossible for the skier to move along the sphere... force towards the center at this point cannot be mv^2/r because the normal force can't be negative...

5. Sep 4, 2007

### Enderless

I got it now! Turns out my teacher made an error.

Thanks everyone

6. Sep 4, 2007

### learningphysics

Cool. Looking at your solution now, I see it is exactly right.