Question

PLANAR KINEMATICS OF A RIGID BODY CHAPTER 16 342 The cylinder shown in Fig, 16-14a rolls without slipping on the of a con point A. The cylinder has a clockwise angular velocity ? 15 rad/s at EXAMPLE 16.7 veyor belt which is moving at 2 ft/s. Determine the velocity of the instant shown. 2tSOLUTION I (VECTOR ANALYSIS) Diagram. Since no slipping occurs, point B on the Kinematic cylinder has the same velocity as the conveyor, Fig 16-14b. Also, the angular velocity of the cylinder is known, so we can apply the velocity equation to B, the base point, and A to determine Velocity Equation (vA)i + (va)yj (%),i +-(%),J 2i (-15k) x (-0.5i +0.5j) 2i + 7.50j + 7.50i so that 2 ft/s tb) (vA) 2+7.50-9.50 ft/s va)y 7.50 ft/s Thus u,-V(9.50)2 + (7.507-12.1 ft/s Ans 7.50 9.50 Ans. SOLUTION II (SCALAR ANALYSIS) As an alternative procedure, the scalar components of VA VS+YAS can be obtained directly. From the kinematie diagram showing the relative “circular" motion which produces ws, Fig. i 6-14c. we 0.5 ft cos 45 Relative motion Fig- 16-14 45 Equating the x and namely ore. y components gives the same results as bef g (e). 2+10.6 cos 45° - 9.50 ft/s (UA),-0 + 10.6 sin 45° = 7.50 ft/s

If the radius of this cylinder is .5 why did he take the r from b to a -.05(i) + .5j since the distance from b to a is not the radius?