The figure below shows a thin, uniform bar of length D 1.32 m and mass M0.65 kg

Question

The figure below shows a thin, uniform bar of length D 1.32 m and mass M0.65 kg pivoted at the top. The rod, which is initially at rest, is struck by a particle whose mass is m0.30 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 600, find the speed of the particle before impact. 113.328xm/s Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. You can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing. Draw a sketch showing the bar and particle just before the collision, and when the bar is at its maximum angle

Explanation / Answer

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 - cos?),

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 - cos?)

v = sqrt{2I(m + M)(g)(ycm)(1 - cos?)} / (0.80mD) [1]  

center of mass of particle and rod (relative to top end) will be :
(ycm) = [(D/2)(M) + (0.80D)(m)] / (m + M)
(ycm) = [(1.32 m)(0.65 kg)/2 + (0.80(1.32 m))(0.30 kg)] / (0.30 + 0.65 kg)
(ycm) = 0.76 m

moment of inertia of particle/rod:
I = (1/3)MD² + m(0.80D)²
I = (1/3)(0.65 kg)(1.32 m)² + (0.30 kg)(0.80(1.32 m))²
I = 0.7956 kg-m²

Substituting ycm, I, etc. into [1],
v = sqrt{2I(m + M)(g)(ycm)(1 - cos?)} / (0.80mD)
v = sqrt{2(0.7956 kg-m²)(0.30 + 0.65 kg)(9.81 m/s²)(0.76 m)(1 - cos60°)} / (0.80(0.30 kg)(1.32 m))
v = 7.48 m/s

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