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