Question

Chapter 3, Practice Problem 3/115 The quarter-circular hollow tube of circular cross section starts from rest at time t = 0 and rotates about point o in a horizontal plane with a constant counterclockwise angular acceleration 7 = 2.9 rad/s. At what time t will the 0.57-kg particle P slip relative to the tube? The coefficient of static friction between the particle and the tube is us = 0.89. tes 58 0.79 m Answer: t = the tolerance is +/-2%

Answer 1

Solution in the uploaded images

Solution olim mrw² Sin32 P58 In juc 18. merw2 10-58 = 32 Ö. 2.8 gead/st m= 0.67 o bring Ms= 0.89 All the force acting acting on a particle (P) is shown in EBD. Let at time t the tube 58 my sino man 32° ing 6332 rom FBD. ng Cas 32 Normal force, Nz Sin 32 tmrwles 32 mg@sin3 For Friction force will be acting tangentially & For partich to slip, equating jories along tangential disuction, mrw?sin 32 Jone ang las32 mrw is the centrifugal force due to violation, From figure, + 0 67x0.67x62 C63 32) = 0.79 & sin 56 M = 0.67m putting nalus in above equation 0.67×0.67 w2sin 32 & 4 x No 0.67×9.8x Cas 32 0-238w'+ 0.84x76-6779-84 Sin 32 0.238w2 & 3.097+ 0.339 6 2 3 5.568 a 5.568 5.566-3.087 0.330 +0.287 1.987 read/

using 14- სსა = equation of motion in scotaction for tube, 1.887 Ö= 2.9 red/ Wy = wotot time, t= 1.187 2.9 Particle will blide out t= 0.685 sec. = 0.685 sec

Answer 2

The quarter-circular hollow tube of circular cross section starts from rest at time t 0 and rotates about point O in a horizontal plane with a con￾stant counterclockwise angular acceleration 2 rad/s2 . At what time t will the 0.5-kg particle P slip relative to the tube? The coefficient of static fric￾tion between the particle and the tube is µs 0.80