Question

Problem 1 (20 POINTS) 0.6 m B The wheel rolls without slipping with a constant 10 rad/s angular velocity. (a) mark the instantaneous center of zero velocity on the picture (b) what is the angular acceleration of the wheel (c) draw the velocity vector of the center of the wheel and show its direction (d) calculate the velocity of the center of the wheel (e) calculate the acceleration of the center of the wheel (f) draw the velocity vector at B and show its direction (g) Calculate the magnitude of the velocity of point B

Answer 1

Solution:- a> Instantaneous center of zero velocity : - 0.6 m W = 10 vadis B A Instantaneous center Since the point of contact will have velocity zero at any instant so, the instantaneous center of zero velocity is point A. b> Angular acceleration : dw at Angular acceleration Since the angular velocity W = 10 rad)s [constant) SO) dw =0 Angular acceleration doo c) velocity vector Of center of wheel:- "0.6m uc y = velocity vector of Center. dy velocity of the center of the wheel uc) Wor > uc oc = 10x0.6 = 6 nys Answer CS Scanned with CamScanner

e > Acceleration of the center of the wheel : - linear alieleration of the wheel center can be given as follows: w2r ac 1028016 ac = 60 m/s2 (f) velocity vector of point B :- wr 0.6m TV 45° Vo «В B So the magnitude of the direction A of velocity will be 450 from horizontal at point B as shown in the above diagram. (9) Magnitude of velocity of polnt B UB=Jvc² + (Wr)2 162 + (1000:6)2 = 562+62 VB = 652 m/s 245 VB = 8:485 m/s 24 45 Answer CS Scanned with CamScanner