The figure below is an overhead view of a thin uniform rod of length L0.6 m and

Question

The figure below is an overhead view of a thin uniform rod of length L0.6 m and mass M=15 kg and traveling horizontally at 60.0 rad/s counterclockwise about and axis through its center. A particle of mass m=5 kg and traveling horiontally at speed 30 m/s hits the rod and sticks. The particle's pathh is perpendicular to the rod at the instant of the hit, at a distance d from the rod's center. At what value of d are rod and particle stationry after the hit? (The moment of inertia of a uniform thin rod through the centr is I=(ML^2)/2)

The figure below is an overhead view of a thin uniform rod of length L0.6 m and mass M=15 kg and traveling horizontally at 60.0 rad/s counterclockwise about and axis through its center. A particle of mass m=5 kg and traveling horiontally at speed 30 m/s hits the rod and sticks. The particle's pathh is perpendicular to the rod at the instant of the hit, at a distance d from the rod's center. At what value of d are rod and particle stationry after the hit? (The moment of inertia of a uniform thin rod through the centr is I=(ML^2)/2)

Explanation / Answer

moment of inertia of rod = Irod = Mrod L2 /12 = (15) (0.6)2 /12 = 0.45 kg m2

angular velocity of rod = Wrod = 60 rad/s

Angular momentum of rod = Lrod = Irod Wrod= (0.45) (60) = 27       anticlockwise direction.

mass of particle = Mparticle = 5 kg

linear speed of particle = Vparticle = 30 m/s

distance from axis of rotation = r = d

angular momentum of particle, Lparticle = Mpartcile Vparticle r = 5 x 30 x d = 150 d     anticlockwise direction

initial Total angular momentum, Linitial= Lrod + Lparticle = 27 + (-150d)

final total angular momentum =Lfinal = 0

Using conservation of angular momentum ::

Linitial = Lfinal

27 + (-150d) = 0

d = 27/150

d = 0.18 m

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