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JIM SWAFO
Answers (1)
angular momentum: l=rmvsinQ The rotational inertia on the rod: I=ML^2/12 The parallel-axis theorem: I=Irod+mr^2 Therefore: rmvsinQ=(ML^2/12+mr^2)w v=(5.5)(0.55)^2+(0.003)(0.275)^2(11.0)/(0.275)(0.003)sin(60.0) v=25618.6 m/s
