# Block on a sphere

1. Apr 1, 2007

### anand

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

A small block slides from rest from the top of a frictionless sphere of radius R.How far below the top,x,does it lose contact with the sphere?(Sphere doesn't move)

3. The attempt at a solution

The angle subtended by the block at the centre of the sphere can be taken as theta.If the x axis is considered tangential to the surface and hence y axis in the direction of the normal force,this problem can be solved,when combined with the conservation of energy.

N-mg cos(theta)=-mv^2/R
and
mgR(1-cos(theta))=1/2mv^2

Answer is R/3.

My question is,how do you solve this problem if the axes are taken "upright",instead of being inclined.i.e,the y-axis is vertical and x-axis is horizontal.

2. Apr 2, 2007

### Mentz114

The choice of axes for X and Y won't affect the result, obviously. You can use the same method for the calculation but X, Y and theta will be different.
If you rotate the axes by angle phi, the new X and Y values will be

X' = Xcos(phi) + Ysin(phi)
Y' = -Xsin(phi) + Ycos(phi)
theta' = theta + phi

If you calculate with X' and Y' you should get the same answer whatever phi is.

3. Apr 2, 2007

### anand

Thanks.
But what if I try to resolve the forces along the "upright" x and y axes,i.e,I get

N cos(theta)=mg+mv^2/R cos(theta)

and

N sin(theta)= mv^2/R sin(theta) and the energy equation.

Now,how do I find the theta at which N becomes zero(which is when it loses contact,right?)

4. Apr 2, 2007

### Mentz114

It's the point at which the force along the radius ( centripetal) becomes equal to the radial component of mg ( vertical). Equate the forces and solve for theta.

 I'm not so sure about this, my first attempt looks wrong and I don't have time to pursue it right now.

Last edited: Apr 2, 2007
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