# Conservation of energy of a sliding block

1. Nov 25, 2007

### anap40

1. The problem statement, all variables and given/known data
A small block starts at rest on top of a frictionless hemisphere (like an igloo) of radius 1.7m. Then the block slides down from the top. At what height above the ground (in m) is the block moving fast enough that it leaves the surface of the hemisphere?

2. Relevant equations

mv^2/r = a

mgh = PE

.5mv^2 = KE

3. The attempt at a solution

I think that the thing will leave the surface when the centripetal force is greater that the force of gravity, because the force of gravity is the only force supplying the centripetal force. So,
(mv^2)/r > mg

so v>sqrt(rg)

The velocity after falling a height h is
v=sqrt(2gh)

sqrt(2gh)>sqrt(rg)

2gh>rg
2h>r
h>r/2

so it would have to fall 1.7/2 = .85m

.85m is not the correct answer.

The only other thing I can think of is that the force supplied by gravtiy varies based on the angle that the normal force between the object and the cube changes.

2. Nov 25, 2007

### PhanthomJay

the centripetal force is provided by the algebraic net sum of the component of the block's weight in the radial direction and the normal force in that direction between the block and the curved surface. When the normal force is 0, there is no longer enough of a radial gravitational component to keep the object moving in a circle. You have neglected that component (gsintheta) in your calculations.

3. Nov 25, 2007

### anap40

Is this diagram correct?
http://img172.imageshack.us/img172/475/33347115yw1.png [Broken]

So the equation should be centripetal force + normal force = cos(theta)mg?

Last edited by a moderator: May 3, 2017
4. Nov 26, 2007

### Staff: Mentor

The appropriate equation would be net normal force = mg cos$theta$ - mv2/r, where $theta$ is the angle with respect to the vertical and the radius extending to the contact point of the mass with the sphere. The diagram is more or less correct.

5. Nov 26, 2007

### anap40

OK, thanks. The equation you gave is just a rearrangement of mine, but I think that I was using the angle between the ground and the radius extending out to the contact point of the mass. If i did it that way I guess I would just have to use sin.

I got the correct answer now, thank you.

6. Nov 26, 2007

### Staff: Mentor

Yes, if one uses the angle with respect to horizontal, then one would use sin rather than cos.