test3csol.fall10

test3csol.fall10 - ME 560 Kinematics Class Test 3 Fall...

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1 ME 560 Kinematics. Class Test 3. Fall Semester 2010 Friday, October 29th Name of Student : ________________________________ OPEN BOOK AND OPEN NOTES Problem (50 points). A disk cam with a reciprocating roller follower is to be designed such that the base circle diameter D = 5 cm, the roller follower diameter d = 2 cm, and the follower eccentricity (or offset) 1cm. = ε The cam is rotating with an angular velocity ω 25 rad / s = counterclockwise and a constant angular acceleration 2 α 4rad/s = clockwise. The lift and the output of the follower center, as a function of the cam angle , θ is specified in the table below. Input (degs) Lift L (cm) Output y 0 – 65 + 4 Full-rise Simple Harmonic Motion 66 – 75 0 Dwell 76 – 165 + 3 Full-rise Cycloidal Motion 166 – 180 0 Dwell 181 – 270 3 Full-return Cycloidal Motion 271 – 280 0 Dwell 281 – 345 4 Full-return Simple Harmonic Motion 346 – 360 0 Dwell At the cam angle 60 , = D compute: (i) The first, second, and third-order kinematic coefficients of the lift curve. (ii) The absolute velocity, acceleration, and jerk of the follower center. (iii) The Cartesian coordinates of the point of contact between the cam and follower in the reference frame attached to the center of the base circle and rotating with the cam. (iv) The radius of curvature of the cam surface. (v) The unit normal vector to the cam surface expressed in the moving reference frame. (vi) The pressure angle of the cam. Is your answer for the pressure angle satisfactory? Briefly explain why or why not.
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2 Solution to Class Test 3. (i) At the cam angle 60 θ , = D the follower is undergoing full-rise simple harmonic motion. Therefore, in the range of the input cam angle 06 5 θ =− DD , the displacement of the follower center, see Equation (6.12a), page 291, can be written as 1c o s 2 ⎛⎞ ⎜⎟ ⎝⎠ L y cm πθ β (1a) where is the input angle of the cam in radians, and 65 65 1.134 radians 180 π == = D (1b) Therefore, Equation (1a) can be written as 4 o s 2 65 180 yc m (2a) or as ( ) 21 co s2 .769 m (2b) Differentiating Equation (2a) with respect to the input angle , the first-order kinematic coefficient of the lift curve is 5.538 sin 2.769 / m r a d ′ = (3) Then differentiating Equation (3) with respect to the input angle , the second-order kinematic coefficient of the lift curve is 2 15.337 cos 2.769 / y cm rad ′′ = (4) Finally, differentiating Equation (4) with respect to the input , the third-order kinematic coefficient of the lift curve is 3 42.472 sin 2.769 / y cm rad ′′′=− (5) Given that the cam angle 60 1.047 rads, D the range 65 1.134 rads, o the diameter of the base circle D5 c m , = the diameter of the roller follower d2 c m , = the follower offset (or eccentricity) 1cm, = ε and the total lift L = 4 cm.
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This note was uploaded on 12/27/2011 for the course ME 560 taught by Professor Na during the Fall '10 term at Purdue.

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test3csol.fall10 - ME 560 Kinematics Class Test 3 Fall...

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