Normal force as centripetal force

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 3K views
Telemachus
Messages
820
Reaction score
30

Homework Statement


Hi. I have this problem:

A particle of mass m, rests on top of a frictionless sphere of radius R. Admitting that part from the rest for the position indicated in the figure along a path contained in a vertical plane.
Get the value of the angular coordinate at the instant the body leaves the surface of the sphere.


attachment.php?attachmentid=28963&stc=1&d=1286746921.png


In the first place I thought: [tex]N-mg\cos\theta=0\Rightarrow{N=mg\cos\theta}[/tex]
And then [tex]N=0\Leftrightarrow{mg\cos\theta=0}\Leftrightarrow{\cos\theta=0}\Leftrightarrow{\theta=\frac{\pi}{2}}[/tex]

But now I'm not sure about this. I think that the angle could be before [tex]\theta=\frac{pi}{2}[/tex], because of the inertia. But I don't know how to raise the problem this way.

Bye there.
 

Attachments

  • sphere.PNG
    sphere.PNG
    1.1 KB · Views: 557
Physics news on Phys.org
The first step is finding out what the speed of the particle at an angle theta is. Conservation of energy is the easiest way to find that.

The normal force and the force of gravity must together produce the acceleration of the mass. Draw a picture of the forces involved. You can separate the forces and accelerattions in a component that is perpendicular and one that is tangential to the surface.
 
I haven't worked yet with conservation of energy, but I see what you're trying to tell me. Then just supposing that the normal force is zero when [tex]cos\theta=0[/tex] its wrong, right?

There is a "kinematic" way of doing this? with polar coordinates maybe?

Thanks for posting willem2.