Sol. At the point A at which the particle leaves the circular trajectory, the normal force vanishes, N = 0 (see figure.) thus, we have

mg sin θ = …...(i)

v A is the speed of the particle at the point A. On the other hand, from the principle of conservation of the total mechanical energy, we know that:

= mgR (1 – sin θ ) + …...(ii)

Therefore, + mR = mgR + …. (iii)

or, v A = …...(iv)

From Eq. (i) and Eq. (iv), we obtain:

sin θ = = = …...(v)

(i) If < , that sin θ < 1, and therefore the particle moves along the circular path before leaving it. Therefore,

h = R sin θ = + …...(vi)

(ii) In the case > , the particle does not even touch the circular path. In this case,

h = R …...(vii)

The general formula of the trajectory of a particle thrown with a initial velocity v 0 and elevation angle θ 0 is (see Figure).

y = y 0 + x tan θ – …...(viii)

In the first case,

y 0 = h = + …...(ix)

and, θ 0 = – (90º – θ ) = –cos –1 …...(x)

Therefore, cos θ 0 = + , and tan θ 0 = …...(xi)

v 0 = v A = …...(xii)

When the particle reaches the ground, y = 0; thus,

0 = + – x – x 2 …...(xiii)

By solving the quadratic equation, we obtain :

x = …...(xiv)

The distance d is (see Figure) :

d = x – R (1 – cos θ ) = ( – )

– R (1 – ) …...(xv)

In the second case, …...(xvi)

Then, we have : 0 = R – ….(xvii)

Therefore, x = –R, and

d = x – R = –R …. (xviii)

One may verify that for v 2 = gR the solutions correspond to each other.