Question

Two blocks are connected by a light rope passing over a pulley of 0.15 m radius and moment of inertia I. The blocks move to the right with an acceleration of 1 m / son ramps with coefficients of kinetic fiction 0.1 between the blocks and the inclined plane a 1.00 m 2 m = 80 kg mg = 10.0 kg 61 32 a) Find the net torque (Nm) acting on the pulley b) Determine its moment of inertia (kgm^2)

Answer 1

May Free body diangarm af blook B. FTB F Fa=26/3² Blog finft toglasi 107 61 l Here, F is the friction force FIB is the tension in the rope N is the normal force on the block. Force in vertical direction N= fog XC6861 x 608 61' = 10x9.81% 60861 N = 47.56 Nt

Colculate the tension in the rope B. Yog sinfi - FTB-Ff = mga M=0.1 a = 1 m/s2 10x9.81 X Sin61 - FTA-MN = 10x 1 85.8 - Fte-01X47:56 = 10 FTB = 71.04 NE Free body diogram of block A, a = 1 m/s2 FT 896832 gesinge Ff 89 32° Here, Fj and FTA is the friction force and tension in the rope. Forte in vertical direction - N= 83 6.1 3.2 32 = 8X9-81X32 M - 66.55 Nt.

Calculate tension in the ope A, FA - 898 n 32 - Fq = maa 24-01 FA-8rg.81% Sinag -UN=801 a = 1 m/s2 FA-47:59 - 0:1X66.55 = 8 IFTA = 56.25 Nt (a) Net Torque pulley = 0.15m That = FIXO FAX8 = 48104-56-25)X0:15 That 2.22 Nom Ana (b) Moment chentia - og Calculente cycle acceleration r 0.15 - 6.67 rad 2 Moment of intritic is I - 2.92 6.67 I = 0.34 kg. m² AM