How to Solve for Radial Component in Rigid Body Rail Problem?

[transgalactic]

here is the question
and how i tried to solve it

http://img242.imageshack.us/img242/7320/83882436ug7.th.gif

 

[chrisk]

This is a frame of reference type of problem. Yes, the marble will release from the track when the normal force is zero. In order for the marble to travel in a circular path a centripetal acceleration must be present. If the observer is the marble, it experiences an inertial force called the centrifugal force. This is the reactionary force caused by the track. So, in this reference frame the centrifugal force is the same magnitude as the centripetal force (v²/R) but in the opposite direction. Therefore, when the gravitational force equals the centrifugal force this is the point where the marble will leave the track.

 

[transgalactic]

i described the equation
can you continue them?

 

[chrisk]

I'm sorry. The web page with your image will not load at this time so I cannot view your equations.

 

[transgalactic]

can you see this one?

http://img242.imageshack.us/img242/7320/83882436ug7.gif

 

[chrisk]

Yes, I can see your image. Thank you. Now, express omega in terms of v (omega = v/R). This will give you an expression for v as a function of alpha. Use the expression for centripetal force in terms of v and equate to the radial component of the graviational force acting on the marble.

 

[transgalactic]

ok what now
the total energy is
$<br /> <br /> & {\rm{mg2R = mgR(1 + cos}}\alpha {\rm{) + }}{{mv^2 } \over 2} + {{I({v \over r})^2 } \over 2} \cr <br /> & \cr}$

 

[chrisk]

Your equation can be solved for v². Now,

$$F_{centripetal}=m\frac{v^2}{R}$$

Find the radial component of the gravitational force as a function of alpha and set it equal to the centripetal force.