A uniform thin bar AB of length L - 6 ft is released from rest at an a.docx

3 de Feb de 2023
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A uniform thin bar AB of length L - 6 ft is released from rest at an a.docx

• 1. A uniform thin bar AB of length L = 6 ft is released from rest at an angle theta = theta1. As the bar slides, the cnds.4 and B maintain contact with the surfaces on which they slide. Neglecting friction and knowing that the cndL4 has a speed of 19 ft/s right before hitting the floor, determine theta1. Solution the instant axis of rotation is the point of intersection of lines drawn from A and B and parallel to the floor and the wall respectively.Let this point be O. Now at theta torque about htis point=weight*(horizontal distance of O from centre of mass of rod) =mg*L{sin0}/2 this should be equal to (moment of inertia)*angular accleration Now, IO=ICM+m(OC)2=mL2/12+m*L2/4=mL2/3 thus mg*L{sin0}/2=mL2/3*dw/dt [where w=angular velocity of rod] dw/dt=3gsin0/(2L) applying chain rule, dw/dt=dw/d0*d0/dt [0=angular displacement] =w*dw/d0 w*dw/d0=(3g/2L)sin0 w*dw=(3g/2L)sin0d0 integrating, we get w2/2=(-3g/2L)cos0+c now at 0=01, w=0 which gives c=(3g/2L)cos0 thus w=sqrt[(3g/L)(cos01-cos0)] now at 0=pi/2 w=sqrt[(3g/L)cos01] and velocity of A=wxr=L*sqrt[(3g/L)cos01] thus L*sqrt[(3g/L)cos01]=19 L2*(3g/L)cos01=361 cos01=361/(3gL)=0.62 thus, 01=51.43 degrees