The spool has a mass of 50 kg and a radius of gyration of ko = 0.280 m. if the 20-kg block a is released from rest, determine the velocity of the block when it descends 0.5 m.
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0 V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J The kinetic energy because the pool rotates about a fixed axis W = VA/rA = VA/0.2 5VA Mass momen of inertila about fixed axis which passes through point 0 I₀ = mko² = 50(0.280)² = 3.92 kg. m² ∴ The kinetic energy of the system is T = 1/2 I₀w² + 1/2mAVA²
= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA² Now that the system is at rest then T₁ = 0 Energy conservation is T₁ +V₁ = T₂ + V₂ 0+ 0 = 59VA² + (-98.1) VA = 1.289 m/s = 1.29 m/s
