# Small Body on a Sphere loss of contact

1. Nov 16, 2012

### jill2040

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

A small body is set on the surface of a smooth sphere at an angle of 45 degrees from the center. At this point the sphere is given a constant acceleration in the horizontal direction of 9.8 m/s/s. There is more in the question, but the goal is to find when the body will leave the circle.

2. Relevant equations

I'm wondering if from the s' frame I can treat the accelerations from psuedoforce due to the sphere's acceleration and the force from gravity as one acceleration toward the center of the sphere for the first quadrant. Since both equal 9.8 m/s/s, I want to simplify dealing with the normal force with a centripetal acceleration of 9.8 m/s/s.

3. The attempt at a solution

I figure since where ever the body lies on that quadrant, the force from gravity will be the sin and the psuedoforce from the acceleration will be the cos of the same angle. Thus, the net force from the two would be
=sqrt( {9.8sinx}^2 + {9.8cosx)^2 )
=sqrt( 9.8^2 {sin^2x + cos^2c} ) and the cos and sin squares would equal 1
=sqrt ( 9.8^2 ) which just equals 9.8

Does this make sense or am I missing something?

2. Nov 16, 2012

### tiny-tim

hi jill2040!
[STRIKE]looks fine to me![/STRIKE]

though a simpler way of putting it would be to say that the two "forces" are equal and perpendicular, so their resultant is at 45°

EDIT: oops, no, it's not fine, i answered too quickly

i think you intended to find the component of each force in the tangential direction, and add them …

in that case, they're the same component, and you just add them (sin45° + cos45°)

Last edited: Nov 16, 2012
3. Nov 16, 2012

### jill2040

So as the sphere accelerates horizontally (and is not rotating I believe) the normal force as the body moves along the surface in the first quadrant is not just equal to m*9.8 m/s/s? I ask because the body starts at 45° and the question wants to know when it would leave the surface.

4. Nov 16, 2012

### tiny-tim

in the reference frame of the sphere (ie, in which the sphere is stationary), the effective gravity is the resultant of g downward and g horizontally …

a total magnitude of … in the … direction?

5. Nov 16, 2012

### jill2040

Ah, the effective gravity won't do what I'm describing, it will only appear that way at one point, which is where I probably got my notion.

6. Nov 16, 2012

### tiny-tim

no, effective gravity is valid for the whole of space! (in the frame of the sphere)