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 0 = 1.1 rad/s2. 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 a = 0.69 0.83 m 53 w Answer: ta S the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work LINK TO TEXT Question Attempts: 0 of 3 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

Azimutbal trajectare is N $ = 37° S = 90°-53 of tube 53 direction, Radial F = MgN' Lire ction Normal fonce Mg = 0.69 m= 0.57 kg 2= Ö= lograd/s? er=0.83m friction force

The normal component using the relation is written as N'= (N+ (mg)? N'=VN+ (0.57*9.818 N'= VN'T 31.267 the weight of the body and normal component are perpendicular to each other. For the radial and azimutal coordinate direction is F =mesa N cosay - F sin 37° mosa W C0537' - MSN Sin 37° = 0.57% 0.83 X 11 N COS 37 – 0.69 VW+31.267 Sin 37° = 0.52041 -0.41525 V + 31.267 0.52041-NC05379 W+31.267 0.52001- N COS37° -0.41525 1.25324 + N(1.92326) N'+ 31.267 (-1.25 324 + N(1.92326 2326)] +31.267 = (-1.252240 +4.923265% 2(1.25324)(1.92326N) N' + 31.267 Zz

=0 -2.6989 NY + 4.8206 N + 99.69638 Solve above expressions we get N= 4.328N (an --96017 The force along the radial coordinate direction is Written as Fay=mags ansin 37 _MG N' cos37 cameror -N Sin 37º – 0.670 +31.267 C0537' -fmos.com) -4.328x Sin 37 – 0.616.328)+31.267 CoS 37° =-0.57 40.83 x(1-11)? -6.50117228 =-0.572451 % 3.3699 sec