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A child sits down on one end of a horizontal seesaw of negligible weight, 2.14 m from the pivot point. No one balances on the o...Asked by john hide
A child sits down on one end of a horizontal seesaw of negligible weight, 1.96 m from the pivot point. No one balances on the other side. Find the instantaneous angular acceleration of the see-saw and the tangential linear acceleration of the child.
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Answered by
bobpursley
wonder if the entire seesaw had any weight, other than no weight on that one end.
If the mass of the child is the only mass in motion, moment of Inertia= mass*1.96^2
torque=I*angular accleration
torque= mass*g*1.96
acceleration= torque/momentofInertion= 9.8/1.96 radians/sec^2
If the mass of the child is the only mass in motion, moment of Inertia= mass*1.96^2
torque=I*angular accleration
torque= mass*g*1.96
acceleration= torque/momentofInertion= 9.8/1.96 radians/sec^2
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