A point object with mass 6 kg and a uniform rigid rod with mass 6 kg and length 11 meter are on a horizontal frictionless planar surface. Point object hits the rod vertically with velocity 8 m/s and sticks to the rod.
Calculate the angular velocity ω about the center of mass just after the point object with mass m sticks to the rod.
Calculate the ratio of the lost energy to the initial kinetic energy of the system?
mass of object , m1 = 6 kg
the mass of rod , m2 = 6 kg
l = 11 m
initial vertical velocity , u = 8 m/s
when the mass sticks to the rod
the center of mass of the system is (L/4) from the center of rod
using conservation of angular momentum
Li = Lf
(m * u * L/4) = (m * (L/4)^2 + m * L^2 /12 + m *(L/4)^2) * w
(8 * 11/4) = ((11/4)^2 + 11^2 /12 + (11/4)^2) * w
solving for w
w = 0.87 rad/s
the new angular velocity is 0.87 rad/s
the ratio of the lost energy to the initial kinetic energy of the system, R = (KEi - KEf)/KEi
R = (0.5 * m * u^2 - 0.5 * (m * (L/4)^2 + m * L^2 /12 + m *(L/4)^2) * w^2 ) /(0.5 * m * u^2)
R = (8^2 - ( (11/4)^2 + 11^2 /12 + (11/4)^2) * 0.87^2 ) /(8^2)
R = 0.7
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