Question

In the figure here, a solid brass ball of mass 0.278 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 0.16 m, and the ball has radius r << R.
(a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop?
(b) If the ball is released at height h = 4R, what is the magnitude of the horizontal force component acting on the ball at point Q?

Answer 1

When the ball is on the verge of leaving the track contact forces act on it which result the ball to leave the track , let us conisder as N

So N+mg(gravational force)=ma(accelartion force acting on the ball)

we know that a=v2/R where v=velocity of ball when it is onverge of leaving the track and R=radius of ring

so N+mg=mv2/R or N=m(V2/R-g)

For fulling the above equation N should not be zero so V=gR

now from the energy conversion principle

mgh(potential engery at the top before release)=mg2R(potenial enegy at the top of ring)+mV2/2(kinetic energy at the track)

so

mgh=mg2R+mV2/2

so mgh=mg2R+mgR/2

h=5R/2=5*.16/2=4 m

(b) as stated above

potentail energy at heught=potenial energy at given point + Kinetic energy at given point

mgh=mg2R+mv2/2 h=4R

mg4R=2mgR+mv2/2

so v=4gR

as stated above in the equation

horizontal force=ma-mg=mv2/R-mg

=m*4gR/R-mg=4mg-mg=3mg=3*.278*9.8=8.17*10^-3 N