# Help please more energy questions

1. Jan 7, 2010

### holezch

1. The problem statement, all variables and given/known data
A point mass m starts from rest (at the top of a sphere) and slides down the surface of the frictionless solid sphere of radius r.

Find the angle at which the mass flies off the sphere.

2. Relevant equations

3. The attempt at a solution

Well, all I know is that at that point, the Normal force will be zero. Also, that it is moving in a circular way.. so that we will have mv^2/r for its radial components and some other tangential components..

please help.. thanks

2. Jan 7, 2010

### rock.freak667

Start by drawing a free body diagram at an angle θ

3. Jan 7, 2010

### holezch

I have a free body diagram, I don't know what equations to use :S

thanks

4. Jan 7, 2010

### ideasrule

Now, write out Newton's second law for the radial direction.

5. Jan 7, 2010

### holezch

okay, so m v^2/ r = mgsintheta - N? When it falls off, N will be 0

thanks

6. Jan 7, 2010

### rock.freak667

Yes so now find what mv2/r works out to be using conservation of energy.

7. Jan 7, 2010

### holezch

hmm.. so should I express v in terms of kinetic energy and stuff? like from 1/2mv^2.. et c

thanks

8. Jan 7, 2010

### rock.freak667

yes that will help.

9. Jan 7, 2010

### holezch

okay, so I got -mgr(1-cos theta) as the change in PE in this case.. I'm having trouble expressing the KE though.. any hints? I need to express the velocity as the function of the position.. the angle..

thanks

10. Jan 7, 2010

### rock.freak667

mgr(1-cosθ) = kinetic energy at the angle θ

So you can find what mv2 is equal to.

11. Jan 7, 2010

### holezch

ah right, of course.. do you mean by the conversation of energy, K = mgr(1-cos theta) at some point?

Thanks! (rats, I should've known)

By the way, thanks for replying back constantly :) there's nothing better than real-time help

12. Jan 7, 2010

### rock.freak667

Well the change in potential energy= change in kinetic energy. Initially there is no kinetic energy so just the change in potential energy is 1/2mv2

13. Jan 7, 2010

### holezch

ah, I see so:

change in KE = - change in PE..

there is no initial KE, so 1/2 mv^2 = - change in PE

but - change in PE = mgr(1-cos theta)

so 1/2mv^2 = mgr(1-cos theta)

:) woohoo

14. Jan 8, 2010

### holezch

awesome, I solved it.. just used mv^2/r and plugged in v from here 1/2mv^2 = mgr(1-cos theta) ...et c :)

Oh.. I had a question, if the tangential part of a force overcomes the radial force, will the object fling out of the orbit?

Thanks

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