Force problem?

Question

I do not need answers I just need some direction here.

A particle of mass m slides without friction on a track. AB is a straight section. BCD is a circular section of radius R. The particle starts from rest at A and leaves the track at point D. [Neglect air resistance].

(a) what force of constraint does the track exert on m at point D just before m leaves the track?

(b) how high does m rise above the ground after it leaves the track?

http:\\www.geocities.com\trunks11111\final1.jpg

Answer 1

a)
You must find kintetic energy of the particle
KE = 1/2 mv².
Use conservation of energy. Vertical distance betwen points A and D is
ΔH = 2R + Rsin30 = 5/2 R.

Once you produce expression for mv² you will be able to find centripetal acceleration of the particle:
a = v²/R

Write dynamic equation in radial direction:
ma = F - mg sin30
and finally derive expression for the force of constraint F:
F = mv²/R + 1/2 mg = 2 KE /R + 1/2 mg

Answer 2

The physics is this. Centrifugal force C = mw^2R = mv^2/R = F; where F is the force you are looking for. w is the particle's angular speed while in the turn, R is turn radius, and v is tangential speed. If C > F; so that C - F > 0 = ma = f, the particle will leave the track with radial acceleration a > 0. By the way, F is called the centripetal force...where did you get the "force of constraint" term?

Since you specify a frictionless environment, F has to come from the push back on the side of the track (on its embankment). And that depends on some embankment angle theta, which was not specified. Further, that same angle would also be the launch angle as the particle left the track. And that is needed to determine H, the max height the particle reaches after it leaves the track. [NB: If theta were zero, H would be zero because the particle would just fly off in a horizontal direction.]