(3) The 12-lb lever OA with 10-in. radius ofgyration about point O is initially at rest inthe vertical position (0 = 90°), where theattached spring of stiffness k= 3 lb/in isunstretched. Calculate the constant momentM applied to the lever at O which will givethe lever an angular velocity o = 4 rad/secas the lever rotates to the horizontal positionat 0 = 0.k = 3 lb/in.ww15"15"

Given: W=12 lb θ=90° K=3 lb/in ω=4 rad/s the radius of gyration k= 10 in h=8 x2=30-152=8.7 inFormula used: U1-2=Mθ ∆Vg=Wh ∆Vg=12kx22-x12 ∆T=12Ioω2 U1-2=work done on the lever ∆T=change in temperature ∆V=change in kinetic energy ω=angular velocity in the lever Work was done on the lever, U1-2=Mθ U1-2=Mπ2 Change in potential energy, ∆Vg=Wh Where W=12 lb, h=8 ∆Vg=12(0-8) ∆Vg=-96 lb.in Change in kinetic energy of spring, ∆Vg=12kx22-x12 Where K=3 lb/in,x2=8.786,x1=0 ∆Vg=1238.7862-02 On solving, ∆Vg=115.6 lb.in Change in kinetic energy on the lever, ∆T=12Ioω2 I=mk2 Where m=1232.2, the radius of gyration k=10, ω=4 rad/s ∆T=121232.2×102×4 ∆T=298.13 lb.inApply work energy, Mπ2=24.8-96+115.6 On solving, M=28.26 lb.in

Answered: (3) The 12-lb lever OA with 10-in.…