Figure 12-11 shows a thin, uniform rod of length 0.600 m and mass M that is rota
ID: 1342418 • Letter: F
Question
Figure 12-11 shows a thin, uniform rod of length 0.600 m and mass M that is rotating horizontally counterclockwise about an axis through its center, at an angular velocity of 76.0 rad/s. A particle of mass M/3.00 and speed 41.0 m/s hits and then sticks to the rod. The particle's path is perpendicular to the rod at the instant of the hit, at a distance d from the rod's center.
(a) At what value of d are the rod and particle stationary after the hit?
m(b) In which direction do the rod and particle rotate if d has a lesser value?
Explanation / Answer
initial angular momentum of the rod = L1 = I1*w1 = +(1/12)*M*L^2*w1
iniital angualr momentum of particle = L2 = -(M/3)*v*d
total iniital angular momentum = Li = L1 + L2
Li = (1/12)*M*L^2 - (M/3)*v*d
after collision the thr rod and particle comes to rest
final angular momentum = 0
from momentum conservation
Lf = Li
(M/3)*v*d = (1/12)*M*L^2*w1
1/3*41*d = (1/12)*0.6^2*76
d = 0.167 m <<-----answer
+++++
if d < 0.167 m
the momentum of rod is more than the momentum of particle
the rod and particle system rotates in counter clock wise
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