What Angle Causes a Mass to Leave a Frictionless Sphere?

[holezch]

Homework Statement

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.

Homework Equations

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

 

[rock.freak667]

Start by drawing a free body diagram at an angle θ

 

[holezch]

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

thanks

 

[ideasrule]

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

 

[holezch]

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

thanks

 

[rock.freak667]

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

thanks

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

 

[holezch]

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

thanks

 

[rock.freak667]

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

thanks

yes that will help.

 

[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

 

[rock.freak667]

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

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

So you can find what mv² is equal to.

 

[holezch]

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

So you can find what mv² is equal to.

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

 

[rock.freak667]

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

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

 

[holezch]

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

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

 

[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