# Energy block physics problem

1. Oct 31, 2006

### ductape

Hey everyone, i have one problem left to do on an assignment and it's driving me crazy, any help would be great.
" A block starts from rest and slides without friction along the surface of a hemisphere of radius R= 3.00 m. As the block slides, eventually it loses contact with the hemisphere. (a) When the block is at an angle of 13.9 deg, what is the speed of the block? (b)What is the angle when the block loses contact with the hemisphere?

2. Oct 31, 2006

### ductape

I might add that for a I tried getting the potential energy at the top, then using trig to find the height at 13.9 degress, and then using conservation of energy to get the new speed, but that doesn't seem to work. This is driving me crazy!

3. Oct 31, 2006

### quasar987

Warning: I suck at mechanics.

With that in mind, how about this: Solve the equation of motion where you suppose that the block is always in contact with the sphere, i.e. where there is always the normal force present. With the coordinate system at the center of the sphere and the initial position of the block at (0,0,R), find the theta dependance of the normal force:

$$\vec{N}=N(\theta)\hat{r}$$

So as I said, solve the equation of motion, and solve for b) first. For b, the idea is that you know that the moment the object leaves the surface is the moment its velocity vector is no longer tangent (or below the tangent) with the surface.

Now that you know at which angle the block leaves the surface, you know if it is still on the surface or not at 13.9 deg. If it is still on the surface, then its motion is still governed by our artificial equation of motion. If not, then just solve for the free falling body problem with initial conditions matching the state of the block when it left the surface.

4. Oct 31, 2006

### PhanthomJay

I think Quasar's approach is correct, but i suck at vector equations. I'd recognize that when the block leaves the sphere, there is no longer any normal force acting, just gravity force mgcostheta toward the center of the circle, and mgsintheta tangent to the circle. Use the centripetal force equation mgcostheta = mv^2/R, coupled with the conservation of energy equation mv^2/2 = mgh (using whatever h works out to be from the geometry as a function of r and theta), and solve the 2 equations with 2 unknowns for theta, no? Any angle above that and the gravity component cannot supply the centripetal force necessary to keep it moving in the circle.

5. Nov 1, 2006

### ductape

ah thanks guys turns out i was doing the first part right, just rounding too much so my answer was thrown off! I took your advice on the 2nd part and it worked out perfectly! Thanks!