Question

2. Two forces F-25 N and F-10 N are applied to the rim of a solid cylinder of radius R -0.2 m. The cylinder starts to rotate from rest, and reaches angular speed of 500 rpm in 5 seconds. (a) Find the magnitude of net torque created by forces F, and F2; (b) Find the magnitude of angular acceleration of the cylinder; (c) Find the mass of the cylinder.

Answer 1

force. ay Magnitude of the Net torque created by F tries to rotate it counter clockwise and Fa also t rotate M Counter clockwise also the angle between porce and radius vector is go. torque due to fi= 8 h = ² x 7 = R = RF, sino o = 90° singo = (counter clockwise) torque due to Fi = RFI Simileriy torque due to F2 = 8 x 2 = Rh Ccounter clockwire) Net torque due to i and B = (RFARE) counter clock wire tur = (25+10) x 0.2 I Net torque = (Fitz) R 17-ON-m Angular acceleration of the cylinder- Let the angular acceleration of the coplinser = x from equation of motion - (Fi + Fa) RC= 140 (met torques IX) heter I is the moment of Inertia of cylinder. ILE 7 N-m or IL-7N Also, giren Hat cylinder starts to rotate from rest- it means Initial ang ular relocity, woo Wo = 0 t =55 w = Боотр» a 500 x 27 rad/s 60 W = 52.33 rad/S

Using equation wa Wotoft W = 52.33 rad/sec Wo = 0 t 5 Sec 52.33 = 0 + 50% x = 5233 rad/ sich V: 10.47 radiaca Magnitude of the angular acceleration of fe cylinder © Mass g 44 cylinder - from equation, (F1+F2) R = IX 35%0.10= Id 7= Id 35X0:20 = I - kgm" I = 7 10.47 I = 0.6687 kg-m² IMR=I that moment of Inertia gte cylinder we know I= I MR? M = 21 " R? M = 2xo. 6687 kig (0.22 M = 33.435kg