Mar 28, 2011 · The figure below shows a thin, uniform bar of length D = 1.21 m and mass M = 0.87 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 ... The angular momentum about point O is deﬁned as the “moment” of the particle’s linear momentum, L, about O. Thus, the particle’s angular momentum is given by, HO = r ×mv = r ×L . (1) The units for the angular momentum are kg·m2/s in the SI system, and slug·ft2/s in the English system.
The velocity of the block just before hitting the rod is Ei = E f ⇒ mgh = (1/2)mv 2 ⇒ v = √(2gh) = 4.43 m/s During the collision, the angular momentum of the system is conserved. The rod and block rotate together after the collision. The total rotational inertia of the system after the collision is Itot = I ro + I bl = (1/3)ML The rotational inertia of a uniform thin rod about its end is ML 2 /3, where M is the mass and L is the length. Such a rod is hung vertically from one end and set into small amplitude oscillation. If L = 1.0 m this rod will have the same period as a simple pendulum of length:
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