Question

A motor is connected directly to an inertial load with viscous damping. The parameters of the system are given as: Jeq = 0.6 kg m² (combined equivalent inertia) JL = 0.4 kg m² (load inertia) B=0.05 Nms (damping) Tem = 2.4 Nm (applied electromechanical torque) a) What is the motor inertia? 2.4 i) 0.6 +0.4 ii) — 0.6+0.4 iii) 0.6-0.4 iv) 70.6 -0.42 b) What is the stead state speed? i) 2.4 -0.05-0.6 ii) 1. 2.4 iii) 2 0.052 2.4.0.05 0.6 iv) 2.4 0.05 c) What is the initial acceleration rate? i)_2.4 ii) 24 0.6+0.4 24 ii) 2.4 iv) 2.4.0.6 0.6 0.05-0.4 0.05

Answer 1

Solution: m . Jeq = JLt JM => JM = Jeg -JL -(0.6-0.4) Cophien üc) (b) Jedê + B88. Et * 3²Jeq+ BS = T where Got displacenerlo Now, wa) = 0u) we speed » O'lm) = 1 ts or Jeq (Swe), + Beau) .ws) - BeatB 1. TIS) s²Jq+ SB ws TV) sy w ). L Ts) s Jeq t B wis) =2.4/5 554 B şcarsteady state speed * lim swis) = x 2.4/5 CamScanher '-TLS) => wy)I - 5 Jeg +B . 'lim oto.05 = 11) 0.05

(o acceleration a6205206) :: Ols) - 1 Tus S²JeqtBs (Multiplying szon both side) % 520") - s2 Tul s²Jeg +Bs * all= TG) 1 Jeq + B/s *a (8) = 2.4 (Dividings? foom numerar and denominats). (as ↑s) = 2-4) s Jeq+B/s initial acceleration rate a lim s als) € 2.4.5 lims srW Sao Jeq+B/s 2-4 ob lophon is) CS Scanned with CamScanner