Say you place a mass of m=10 kg on one side of a see-saw, like shown. A child of

Question

Say you place a mass of m=10 kg on one side of a see-saw, like shown. A child of mass M = 25 kg sits on the other side. The height is h = 0.3 m, and r = 1.5 m. The see-saw tips down on the child's side, accelerating until the child hits the ground. Assume that the force of gravity exerts a force perpendicular to the position vector, so phi = 90^degree always. The mass is launched after the child hits the ground, going straight up, like shown. For the moments of inertia treat each of the objects as point masses and treat the see-saw itself as massless, so the total moment of inertia for the system is I = mr^2 +M(3r)^2 in the configuration shown. a) What is the maximum height that the mass reaches?

Explanation / Answer

angle turned= arcsin h/r + arcsin h/3r

= arcsin 0.3/1.5 + arcsin 0.3/4.5

= 0.268 rad

angular acceleration = torque/i

= (25*9.8*4.5 - 10*9.8*1.5)/(10*1.5^2+25*4.5^2)

= 1.807 rad/s^2

by third equation of rotation, w = sqrt(2 alpha theta)

= sqrt(2*0.268*1.807)

= 0.984 rad/s

maximum height = 0.3*4/3 + v^2/2g

= 0.4 + (0.984*1.5)^2/(2*9.8)

= 0.511 m answer

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