• Author / Uploaded
• Gabriela Sequeira

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PROBLEM 15.1 The motion of a cam is defined by the relation 9 = 4f3 - 12r2 +15, where 9 is expressed in radians and t in seconds. Determine the angular coordinate, the angular velocity, and the angular acceleration of the cam when (a) t = 0, (b) t = 6 s.

SOLUTION 9 = (4t3 - 1212 + 15jradians

a) - — = (l2r2 - 24r)rad/s dt v >

(a) t = 0,

= (241 - 24) rad/s2

9 = 0 + 0 + 15 (o = 0 + 0 a = 0 - 24

(b) t = 6 s,

9 = (4)(6)3 - (12)(6)2 +15 = 447 a> = (12)(6)2 - ( 2 4 ) ( 6 ) = 288 a - (24)(6) - 24 = 120

9 = 15.00 radians A S il o A

a =

a = -24.0 rad/s2 M 9 = 447 rad < co = 288 rad/s -4 a = 120.0 rad/s2 4

PROBLEM 15.2 For the cam in Prob. 15.1, determine the time, angular coordinate, and angular acceleration when the angular velocity is zero.

SOLUTION 6 = ^413 - 1212 + 15jradians

03

- — = ( l i t 2 - 24f)rad/s dt ' >

a =

- (24r - 24)rad/s2

Set 0) - 0, and solve the quadratic equation for t. The roots are t =0 At t = 0,

At t = 2 s,

and

t =2s

0 and 2.00 s -4

0 - 0 + 0 + 15

6 - 15.00 radians 4

a = 0 - 24

a = -24.0 rad/s2 4

0 = (4)(2)3 - ( l 2 ) ( 2 ) 2 +15 = - l a = (24)(2) - 24 = 24

6 = -1.000 radian 4 a - 24.0 rad/s2 4

PROBLEM 15.3 The motion of an oscillating flywheel is defined by the relation 9 = 9 0e~3xl cos4;rr, where 9 is expressed in radians and I in seconds. Knowing that# 0 = 0.5 rad, determine the angular coordinate, the angular velocity, and the angular acceleration o f the flywheel when (a) t = 0, (b )t = 0.125 s.

SOLUTION

9 = 0.5 e~3xt cos4nt a) = — = 0.5(-3;re~3" cos4/rf - 4ne~3n' sin4^r/) dt ' > a =

dt

= 0.5(9/r2e 3,rl cosAnt + \2n2e~3xl sm4nt + \2n~e~3xt sin4^/ - 16;r2e 3x1cos4/rt j v

(b) t - 0.125 s,

L*

II

II ©

= 0.5^24/r2