机器人学导论(克雷格)课后习题第四章作业答案

3.2
Solution:
According to the question, we need to calculate the total kinetic energy of the system, including the kinetic energy of the link and the roller.
The mass of the roller is negligible, so the kinetic energy of the roller can be neglected.
The kinetic energy of the link rod includes two parts: one is about the linear motion, and another is about the rotation motion.
Particularly, we can notice that the velocity of the pendulum is the summation of the velocity of the roller and of the link rod relative to the roller, or
/000ˆˆˆˆ*(cos sin )22
cos sin ˆˆˆ22
sin cos ˆˆ()22
C A C A
V V V l l V i k i J l l V i J i l l V i J θθθθθθθθθθθ=+=++=+-=-+ And
222222220022200sin sin cos 44
sin 4
c l l V V V l l V V l θθθθθθθθθ=+⋅-⋅+⋅=+-⋅ When “I ” is moment of inertia of the link rod, we can get that
222222001122
11(sin )242
c K mv I l m V V l I θθθθθ=+=+-⋅+ 3.4
Solution:
For a six-axis cylindrical-RPY robot, joints 1 and 3 are prismatic , joints 2,4,5,6 are revolute.
The constant matrix Q R for revolute joints is as follows:
0100100000000000R Q -⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦
The constant matrix for prismatic joints is as follows:
0000000000010000P Q ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦
According to equation 4.25 in the text book:
006123456;ij i ijk k k j U T U T A A A A A A q q q ⎛⎫∂∂∂=== ⎪ ⎪∂∂∂⎝⎭
Thus , we get:
0662*******
R T U AQ A A A A A θ∂==∂ 03355
0T U θ∂==∂ 0553123456P T U A A Q A A A A l
∂==∂
626231234563
R P U U AQ A Q A A A A θ∂==∂ 535341234564P R U U A A Q A Q A A A θ∂=
=∂ 4.8
Solution:
According to the given conditions:
[]1050500T
R F = ()()()0,1,0;0.707,0,0.707;0.707,0,0.707n o a ==-= ()()()4,6,3;10,5,0;5,0,0p f m ===
According to equations 4.68 in the text book ,we get:
57.077.07o
x o y o z f n f f o f f a f =∙==∙=-=∙= ()20,30,40b f p m =⨯+=-
3014.1442.42
o x o x o x m n b m n b m n b =∙=-=∙==∙= Thus , we get the final result:
[]57.077.073014.1442.42T o F =--。