A uniform rod of length L1 = 1.4 m and mass M = 2.0 kg is supported by a hinge a

Question

A uniform rod of length L1 = 1.4 m and mass M = 2.0 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L2 = 1 m from the hinge. The particle sticks to the rod on contact. After the collision, max = 45°.

Figure 10-52
(a) Find m.
kg
(b) How much energy is dissipated during the collision?
J

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Explanation / Answer

moment of inertia of the rod about the axis. I=ML1^2/3=1.31 energy is conserved so we have. I*w^2/2=M*g*L1/2 so ML1^2*w^2/3=M*g*L1. so L1*w^2=3*g. so that w=sqrt(3g/L1)=4.6(rad/s). ------ angular momentum is conserved. w*I=w'*I' where w' be angular velocity of the system after collide. I'=I+m*l2^2 so 4.6*1.31=w'*(1.31+m) so w=6.03/(1.31+m). energy is conserved. so that. w^2*I'/2=M*g*L1*(1-cos45)/2+m*g*L2*(1-cos45). so that (6.03)^2/2(1.31+m)=4.02+m*2.87. so 18.2=(4.02+2.87m)*(1.31+m). so 18.2=5.3+7.8*m+2.87m^2. solve this we have m=1.2(kg)

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