Question

A thin stick of mass M = 3.1 kg and length L = 1.6 m is hinged at the top. A piece of clay, mass m = 0.7 kg and velocity V = 4.0 m/s hits the stick a distance x = 1.20 m from the hinge and sticks to it.
1. What is the angular velocity of the stick immediately after the collision?

2. What is the ratio of the final mechanical energy to the initial mechanical energy?

Answer 1

angular momentum around pivot is the same before and after

before
angular momentum = m v r = .7*4*1.2
= 3.36 kg m^2/s

after let w = omega = ang velocity about pivot
angular momentum = I w

I = (1/3)(3.1)(1.6^2) + .7(1.2^2)
= 2.645 + 1.008 = 3.65

so
I w = 3.36
3.65 w = 3.36
w = .921 radians/second

part 2
before
(1/2) m v^2 = (1/2).7(16) = 5.6 Joules
after
(1/2) I w^2 = (1/2)(3.65)(.921^2) = 1.55 Joules

1.55/5.6 = .276