The two pendulums shown have 6 kg balls supported by rigid m

The two pendulums shown have 6 kg balls supported by rigid mass-less rods. The pendulums rotate on frictionless pivots in the same vertical plane in a circular path of radius 4 meters and collide. If each is at rest and oriented at theta = 56 degrees then released simultaneously, determine the impulse of the collision in kg-m/s if the coefficient of restitution = 0.8

Solution

lets take pivot point as reference point.

initial vertical position of centre of mass of system is Yi

then Yi = Lsin@ = 4 sin56 = 3.316 m

final when pendulum rod is vertical.

Yf = - 4 m

Using energy conservation to find speed at final position,

initial PE + KE = final PE + KE

mgYi + 0 = mgYf + mv^2 /2

9.8 ( 3.316 - (-4)) = v^2 /2

v = 11.98 m/s

coefficient of restitution = velocity of separation / velocity of approach

0.8 = (11.98 + 11.98) / (vf + vf)

2vf = 23.96 / 0.8

vf = 14.98 m/s

Impulse = change in momentum of ball

= m (vf - v)

vf and v are in opposite direction so.

Impulse = 6 ( 14.98 - (-11.98))

= 161.74 kg . m/s