Circular motion with unknowns and picture

[breck]

Homework Statement

An ice cube slides on a frictionless sphere. It is placed at the top at of the sphere with radius (R), and given an instantaneous push at speed (S). At what angle(x) from vertical does the ice cube leave the surface of the sphere?

Radius = R
Speed/Velocity = v
angle = x

Homework Equations

F=ma
v_0^2=v^2+2gRcosx

The Attempt at a Solution

mgcosx=mv^2/R
Rgcosx=v^2
Rgcosx=v_0^2-2gRcosx
3Rgcosx=v_0^2
cosx=v_0^2/3Rg
x=arccos(v_0^2/3Rg)

this seems like a processed answer but i believe it is wrong. i think step 3 (Rgcosx=v_0^2-2gRcosx) is wrong. But even still, my professor said that that v was instantaneous. so v_0=v?

I honestly don't know how to go about this problem. I tried going about it the same way as a similar problem (mass at the end of a string, what is the angle that T=0) but it does not translate well to this problem. any type of nudge in the right direction would be excellent.

 

[ehild]

How did you get that equation for the angle?

ehild

 

[breck]

How did you get that equation for the angle?

ehild

Like I said, I was trying to follow a similar type problem, but it didn't work out well and that is just a mismatched answer. It is wrong. I just have no idea about how to solve this problem correctly.

 

[kjohnson]

Just like any circular motion problem you need to sum the forces in radial direction and set that equal to mv^2/r. In this case your going to have a component of your gravitational force that can be expressed as a function of the angle along with the normal force.

Now if it is looking for the point when the block leaves the sphere what is the normal force going to be at that point?

Next since there is no friction you can determine the velocity at any point using...

 

[breck]

Normal force would equal mg. mg+Rcosx=mv^2/R?

 

[kjohnson]

No, the normal force would only equal to mg at the very top if it was not in motion. The normal force is the force applied on the block by the sphere. So if they are not touching that force is...

For the other part of your equation you are on the right track with Rcosx, but the 'R' is not correct. Your trying to find the component of the weight in the radial direction. Also Rcosx would not be the correct units for force.

 

[venkatg]

I get theta = arccosine((u^2 + 2gr)/3gr)

 

[breck]

I get theta = arccosine((u^2 + 2gr)/3gr)

that is the answer i get for a similar problem but i don't think it is the answer to this problem
the normal force would be mgcosx. that is correct units