EXAMPLE 4: Newton’s 2^nd law

A particle starts at A with an initial velocity of , and moves down the smooth circular ramp up to point B when it separates. Obtain the equation of motion of the particle and the angle at which it separates.

Solution:

Kinematics:

r = constant

Kinetics:

(2) is the equation of motion

At point of separation, N=0

Integrating (2) using

we get

By the initial conditions at A

Substitution of C into (4) gives

Solving for and substituting into (3) gives