• Document: Solutions to Chapter 8 Exercise Problems
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Solutions to Chapter 8 Exercise Problems Problem 8.1 A cam that is designed for cycloidal motion drives a flat-faced follower. During the rise, the follower displaces 1 in for 180˚ of cam rotation. If the cam angular velocity is constant at 100 rpm, determine the displacement, velocity, and acceleration of the follower at a cam angle of 60˚. Solution: The equation for cycloidal motion is:   y = L   1 sin 2   2   For L = 1, and  = 180˚=  , then     2   (  2   2) ( y = L   1 sin 2 = 1   1 sin 2 =   1 sin2 )   y˙ = L 1  cos 2 =  (1 cos2)     2 2     () y˙˙ = 2L  sin 2 = 2  sin2 The angular velocity is ˙ = 100 rpm = 100 2 = 10.472 rad / s 60 When  = 60˚=  , 3 y= (3  21 sin2( 3)) = (13  21 sin(2 3)) = 0.195 in y˙ =  (1 cos2 ) = 10.472 (1 cos(2 3)) = 5.000 in.   sec. 2 2 () ( ) y˙˙ = 2  sin2 = 2 10.472 sin(2 3) = 60.46 in.2   sec Problem 8.2 A constant-velocity cam is designed for simple harmonic motion. If the flat-faced follower displaces 2 in for 180˚ of cam rotation and the cam angular velocity is 100 rpm, determine the displacement, velocity, and acceleration when the cam angle is 45˚. Solution: - 328 - The equation for simple harmonic motion is:   y = L 1 cos  2  For L = 2, and  = 180˚=  , then   2  2 (  ) y = L 1 cos  = 2 1 cos  = (1 cos ) dy d ( y˙ = = 1 cos ) = ˙ sin dt dt d 2y d ˙ y˙˙ = = ( sin ) = ˙2 cos dt 2 dt The angular velocity is ˙ = 100 rpm = 100 2 = 10.472 rad / s 60 When  = 45˚, y = (1 cos ) = (1 cos45˚) = 1 0.707 = 0.292 in , y˙ = ˙ sin = 10.472 sin45˚= 10.472 (0.707) = 7.405 in s y˙˙ = ˙2 cos = 10.4722 (0.707) = 77.531 in2 s Problem 8.3 A cam drives a radial, knife-edged follower through a 1.5-in rise in 180˚ of cycloidal motion. Give the displacement at 60˚ and 100˚. If this cam is rotating at 200 rpm, what are the velocity (ds/dt) and the acceleration (d2 s/dt2 ) at  = 60˚? Solution: The equation for cycloidal motion is:   y = L   1 sin 2   2   For L = 1.5, and  = 180˚=  , then     2   (  2  ) ( y = L   1 sin 2 = 1.5   1 sin 2 = 1.5   1 sin2  2 )   y˙ = L 1  cos 2 = 1.5 (1 cos2 )     - 329 - 2 2 2      () y˙˙ = 2L  sin 2 = 2(1.5)  sin2 = 3  sin2  () The angular velocity is ˙ = 200 rpm = 200 2 = 20.944 rad / s 60 When  = 60˚=  , 3 3 1 ( 2 ) ( y = 1.5   1 sin2 = 1.5   2 3 ) sin 2 = 0.293 in y˙ = 1.5 (1 cos2 ) =  1.5(20.944)  ( ) 1 cos 2 = 15.00 in 3 s 2 2 ()   ( ) y˙˙ = 3  sin2 = 3 20.944 sin 2 = 362.76 in2 3 s When  = 100˚= 100 = 5 , 180 9 5 9 1 ( 2  ) ( y = 1.5   1 sin2 = 1.5   sin 10 = 0.915 in 2 9 ) Problem 8.4 Draw

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