Mp4006 mp4g02 sem 1 0910

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee

Z (K)

1.

z (k) X (I) x (i) Y (J) y (j) System XYZ is fixed at origin O, and does not rotate with the L-bar System xyz is fixed at origin b, and rotates with the L-bar Assuming the L-bar has rotated 500 cos

200 sin

500 sin 500

0 ,

sin

200

500 cos /

200 cos

500 cos

500 sin

2

500

500 sin

200 cos

cos

500 Since

about the Z-axis (CCW), we have:

sin

200

200 sin 200 cos

cos 3

and

200 sin 3

/

/

. Therefore, we have:

200

400

1000

2600

700

Based on the coordinate systems set up, point P is fixed with respect to system xyz. Therefore, we have: 50 0

50 0

and

As mentioned, system xyz rotates with the L-bar, which means it rotates about the Z-axis. Therefore, we have: 2 K

Angular velocity of system xyz relative to system XYZ = 3 Based on equation 5.40 and 5.45 in the notes, 400

1000

2

50

0

400

mm/s

2

2600

700

3

50

2

2

50

0

2 2

0

2600

700 mm/s

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee 2.(a)

; 0 1 0 0

;

0 0 1 0

1 0 0 0

;

0 1 0 0

0 0 1 0

0

1 0 1

;

0

0 2 0

1 0 1 1

; 1 0 0 0

;

0 2 0 1

(b)(i) Joint

Joint variable -90

o

2

-90

o

3

0

1

(ii)

-90

o

-90

o

0 0

0

0 cos

cos sin 0 0

cos

cos

0

cos sin 0 0

0 0 1 0

cos 0 0

cos sin 0 0 1 0 0 1 0 0 0 0

0 0 1 0

cos 0 0

0 0 1 0

0

1

0 0 1 cos

0 0

0 1

1

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee 0 0 0 1

(iii) 1

0 0 0 1

1

cos sin 0 0

cos 0 0

cos sin cos 0

cos cos sin cos

1

cos cos sin cos sin 0

1

cos cos sin cos sin 0

cos 0 0

cos sin cos 0

cos sin cos

1

0 0 1 0

0 0

cos 0 0

cos sin cos

cos sin 0 0

1

0 0 1 0

cos cos 0 0

1 0 0 0

0 1 1 0 0 0

1

0 1 0 0

0 1 0 0 0 0 1 0

0 0 1 0 0 0 1

cos cos sin cos 1

0 0 1

0 0 0 1

0 0 0 1 0 0 0 1

cos cos sin cos 1

cos sin cos

cos cos sin cos

Differentiating the above, we obtain: cos sin

sin cos cos

sin cos

sin cos cos cos

cos sin sin cos sin

0

sin cos cos cos cos sin sin sin

sin

cos sin sin sin cos sin cos

Therefore, the Jacobian matrix is: sin cos

sin cos cos cos 0

cos sin

cos sin sin sin sin

cos sin cos

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee 3.(i)

Since

and

are velocities relative to the spinning arm, we have the following:

The rotation of disk with centre B with respect to the fixed frame = The rotation of disk with centre C with respect to the fixed frame = Velocity at point B, Velocity at point C, Therefore, we have

Total kinetic energy of the system,

Taking height at point A as the datum, we obtain the following: 0 ·

· · ·

· ·

Therefore, the total potential energy of the system, ·

· ·

(ii) The Lagrangian, L=K-P L=

·

· ·

The required motor torque,

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee 4.(i) A workspace is the set of spatial locations where a robot’s last link can reach.

Consider the following cases: (a) L1 > L2 We have half of a ring shape for Y > 0 with the outer radius of L1 + L2, and an inner radius of L1 – L2. We have one rounded end, defined by a semicircles at (L1, 0) with a radius of L2 and the other side is a flat end, as there is a table present obstructing link 2 from making the full 360o of motion.

Y

X (workspace is shown in yellow in the diagram) (b) L1 = L2 We have a semicircle for Y > 0 with a radius of L1 + L2.

We have one semicircles for Y < 0, centered at (L1, 0) with a radius of L2 and the other side is a flat end, as there is a table present obstructing link 2 from making the full 360o of motion.

Y

X

(workspace is shown in yellow in the diagram) (c) L1 < L2 We have a semicircle for Y > 0 with a radius of L1 + L2.

We have a part of a semicircles for Y < 0, centered at (L1, 0) with a radius of L2 but terminating at X < 0 as there is a table obstructing the motion. The flat end at X < 0 and Y < 0 is due to the presence of a table obstructing link 2 from making the full 360o of motion.

Y

X

(workspace is shown in yellow in the diagram)

(ii) For 1 link of length L, if it has rotated to an angle at 0.1o off from the desired angle, its corresponding position would be off by: e=

· 0.1 ·

Therefore, the combined position error of joint 1 and joint 2 is: E=

·

·

The accuracy of position for the robot’s gripper is DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee (iii) Assuming a triangular velocity profile: Angular velocity, rad/s

0 Total angular displacement = 2 ⁄5

Therefore,

Angular acceleration,

·

2.5

5

Time, s

·5

/ ⁄2.5

/

Assuming the total weight of the robot arm to be a point mass and the two links to be arranged as shown in the diagram: Y 0.30 m

0.20 m 20.0 kg

3.0 kg

X

We obtain the mass moment of inertia about joint 1’s rotation axis as follows:

∑ 20.0 2.55

0.30

3.0

0.50

The maximum power is at the point of peak angular velocity and is obtained as follows:

2.55 1.61

(iv) Propose a curve of 3rd order polynomial:

When

,

, and we obtain,

When

,

, and we obtain,

When

,

, and we obtain,

When

,

, and we obtain,

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee The coefficients inside the proposed equation can be obtained by solving the above 4 equations, or by using the matrix approach as follows:

5.(i) Let

be the rotated angle of the laser-gun CCW from a line parallel to the X-axis on a circle

centered at (0, h).

Therefore, the equations of trajectory are: cos sin (ii) Using a feed-forward control to compute the desired control signals, and using a feedback control to minimize the difference between the desired output and actual output. The system under control in this case is the motors at joint 1 and joint 2.

(iii) Assume k to be the large speed reduction ratio, and ∆

to be the angular displacement at the motor shaft.

As such, the angular displacement at the joint would be given by ∆ / .

The advantages of this situation is: (a) To achieve a certain accuracy of angular displacement (or angular velocity and angular acceleration) at the joint of say 0.1o, the controller only needs to achieve an angular displacement accuracy of 0.1 k o at the motor shaft. (b) The sensors used for the feedback control does not have to be of such high resolution, as it can be placed at the motor shaft.

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.

Brought to you by Mechanical and Aerospace Engineering Club Academic Committee (iv) The following diagram illustrates an incremental encoder:

Counter

∆ (a) We need to know the incremental angle, ∆ , which is the difference in angular position between 2 holes on the disc, as shown in the diagram. (b) A counter is used to count the number of increments at detector A. (c) Detector B is at an offset of ¼ cycle from detector A and is used to determine the direction of rotation (depending on which signal has the 0-to-1 transition first). (d) Detector r is used to determine the reference position, . (e) As such, the final angular position is ∆

DISCLAIMER: The solutions are done by students who scored A or above in this subject. The MAE Club and Campus supplies are not liable or responsible for any errors in the contents of these solutions. Students are advised to take the solutions as a guide rather than absolute answers to exam paper.