Asked by anonymus
A thin stick of mass M = 2.0 kg and length L = 2.6 m is hinged at the top. A piece of clay, mass m = 1.0 kg and velocity V = 0.9 m/s hits the stick a distance x = 1.90 m from the hinge and sticks to itWhat is the ratio of the final mechanical energy to the initial mechanical energy?
Answers
Answered by
Damon
angular momentum is constant
energy goes down
Initial L around hinge = I omega of stick + m v R of clay
I = (1/3) * 2.0 * 2.6^2 * 0 + 1.0 * 0.9 * 1.90 = 1.71 kg m^2/s
Final L = 1.71 = 4.51 omega + 1.0 * 1.9^2 omega= (4.51+.0361)omega
= 8.12 omega
omega = .211 radians/s
initial Ke = (1/2) m v^2 = (1/2)(1)(.81) = .405 Joules
final Ke = I omega^2
I = 4.51 (.211)^2 + .0361 (.211)^2 = .487 (.211^2) = .0217 Joules
.0217/.405 = 0.0536
in other words lost most of the Ke in the collision
energy goes down
Initial L around hinge = I omega of stick + m v R of clay
I = (1/3) * 2.0 * 2.6^2 * 0 + 1.0 * 0.9 * 1.90 = 1.71 kg m^2/s
Final L = 1.71 = 4.51 omega + 1.0 * 1.9^2 omega= (4.51+.0361)omega
= 8.12 omega
omega = .211 radians/s
initial Ke = (1/2) m v^2 = (1/2)(1)(.81) = .405 Joules
final Ke = I omega^2
I = 4.51 (.211)^2 + .0361 (.211)^2 = .487 (.211^2) = .0217 Joules
.0217/.405 = 0.0536
in other words lost most of the Ke in the collision
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