Question

P (N) The 40-kg disk is rotating at o = 100 rad/s. When the force P is applied to the brake as indicated by the graph. If the coefficient of kinetic friction at B is Mk = 0.3, determine the time t needed to stay the disk from rotating. Neglect the thickness of the brake. 500 1 1 t(s) 2 P 300 mm 300 mm B 200 mm 150 mm

Answer 1

disk mass= uokq radius of disko): 150mm 0:15 m rotation (W)- 100 pools. Gefficiend of kinetic friction (M)= 0 3 FBD, ? 0.3 m 0.3 m Nanomal Reaction at dsk-lo injurforumu. 0.3N 0.2 N he hely Ox "Ay foo free body diagram Оy , Фі y fiabe are & An & Ay are Grop Reaction on o & A in ub n sy direction. since slipping occurs at brake pad Ep Up N = 0:3N from figila) - Free body # EMACO, diagram of lever. (0.6) - [ 10.2) - P 10.3)=0 N10.6) - (0:3) N (0:2) - P 10.3)=0 0.54 N 0.3 P=0 0.3 Na p=0.555 P 0:54 thus Ff ( frietion frue) = UN = (0.3) (0.555)P = 0.1667P The mass moment of inertia of disk about its centre o, is To 1 m 2 2 = 1 (40) 10:15)² = 0.45 kq m2

Applying principle of Impulse a momentum, Io w, te Modt= Io Wa - 10 values of Io. Mo, W., W2 in Cadao eqo subtituting E MO= 0.15 Ff 0.45 (0) (045) (100) + Dt -0-1667 P101513dt 45+ Set [-0.1667)(0.15) P] dt=0 p dt sete 45 0.025 = 1800 from sot pat pat diagram, area under (wire represents Stedt, PEND Sco VASIL tisoo)(2) + 500 (+-2) = 1800 2 A t(s) shot 500 (+-2) - 1800 sool+-2) - 1300 13 t-2= 1300 5 t-2-2.6 time te 2.6+2 = 4.6 needed Note a All length dimensions are taken / converted in meter.