Question

A thin, uniform bar of length D = 1.00 m and mass M = 0.74 kg pivoted at the top. The rod, which is initially at rest, is struck by a particle whose mass is m = 0.30 kg at a point x = 0.80d below the pivot. Assume that the particle sticks to the rod. If the maximum angle between the rod and the vertical following the collision is 60°, find the speed of the particle before impact.

Answer 1

angular momentum of particle before hitting rod:
L = mvr, where r = 0.80D
L = 0.80mvD

kinetic energy of particle/rod just after impact = gravitational potential energy when rod is at maximum angle
K = (m + M)(g)(h), where h is the height of the center-of-mass above its initial position
L²/(2I) = (m + M)(g)(ycm)(1 - cosθ), where ycm is the center-of-mass of particle/rod relative to top end, I is the moment of inertia of particle/rod system
(0.80mvD)²/(2I) = (m + M)(g)(ycm)(1 - cosθ)
v = sqrt{2I(m + M)(g)(ycm)(1 - cosθ)} / (0.80mD) [1]

center of mass of particle and rod (relative to top end):
(ycm) = [(D/2)(M) + (0.80D)(m)] / (m + M)
(ycm) = [(1.00 m)(0.74 kg)/2 + (0.80(1.00 m))(0.30 kg)] / (0.30 + 0.74 kg)

(ycm) = 0.586 m

moment of inertia of particle/rod:
I = (1/3)MD² + m(0.80D)²
I = (1/3)(0.74 kg)(1.00 m)² + (0.30 kg)(0.80(1.00 m))²
I = 0.4386 kg-m²

Substituting ycm, I, etc. into [1],
v = sqrt{2I(m + M)(g)(ycm)(1 - cosθ)} / (0.80mD)
v = sqrt{2(0.4386 kg-m²)(0.30 + 0.74 kg)(9.81 m/s²)(0.586 m)(1 - cos60°)} / (0.80(0.30 kg)(1.00 m))
v = 3.305 m/s