Question

The 11.3 kg double pulley shown is at rest and in equilibrium when a constant 4.75 N.m couple M is applied. Neglecting the effect of friction and knowing that the radius of gyration of the double pulley is 152 m, determine (a) the angular acceleration of the double pulley, (b) the tension in each rope. 100 mm 4.5 2.27 kg B

Answer 1

11.3 kg is 4.75 N-m Solution · Moss of the pulley (m) = Couple applied (m) = Radius of Gyration of the pulley (K) = 152 m angular acceleration of . assume the pulley is the direction L of (in rad/ see 2) (ople same in direction. m then. → Linear acceleration of the mass A is ад - V x2 where, r=radius of the pulley Laa = 0.102 xH m/s2 Linear acceleration of the mass B is OB Yo x 2 where Linear acceleration direction is same as angular Ace direction ад =0.203x2 m/s2 assume Tension in rope A is ТА and in rope B is To then m a 9 3 ab CSScanned with CamScanner AA

F. B.D (free Body diagram) of mass Ais : apply Newton's and law i 1 TA A Ha Enet mass x Acceleration ад (mag - Ta) = maxaA Mag 4,5 x9.81 -TA 4.5 x 1•10282) TA 44.15 -0.459 2 -0 F.B.D of mass B is : TB apply Newton's and Law : B fag Fnet mass x Acceleration = MB g (TB - MB xg) mg хав Тв 2.27 x 9.81 2:27 X 10.203X2) TB = 29.269 + 0.46082 2 moment of Inertia Inertia of the double pulley is mass of the pulley X (Radius of gyration) I 11.3 X (152) I = 261075-2 kg-m Scanned with CamScanner

F. B.D of the double pulley pulley is : 8 A O V TB I A : بینها Newton's Apply an and moment about point o moment of Inertia X angular accleration Emnet I L m+ TAXVA - TBX8B = Io a TB X 0.203 4.75 + TAX 0.102 261075.2 X2 13 eq ☺ Value from eq 0 and put in eq ③ them; 4.75 + 44.15-0.4592) X0.102 - (22:27+0.4612)* • 203 = 261075,2 x 2 4.75 -0.142 = 261075.2 x 2 0.000018 rad see? - 2 Scanned with CamScanner

Pulley is (a) Angular Acceleration of the d = 0.000018 rad (seed] Any 1672 value put in eq o TA 44.15 -0.459 x 2 TA = 44,1499 N → 2 value put in eq @ TB - 22.269 +0.4608 xd ITB Scanned with Camscanner 22.2691 V

Note: You will check the unit of radius of ggration. If radius of gyration is 152 mm instead off 152 m then you will comenenta then I will CSScanned with CamScanner solve again if you required.