- 1 -

ENG1040 Engineering Dynamics

Practice Class 10 Instantaneous Centres and Velocity Diagrams - Solutions

Question 1

To calculate the velocity of the slider, we realise that it is attached to member 3, and therefore its instantaneous velocity will be equal to the instantaneous velocity of the pin connecting member 3 to the slider, V4.

To find velocity V4, we first need to know the location of the instantaneous centre of member 3, its angular velocity, and its distance from the slider pin.

To find the location of the instantaneous centre of member 3, find two points on the member at which we know the velocity. These are the velocities of the pin connecting members 2 and 3 (V23), and the pin connecting member 3 to slider 4 (V4). The intersection of lines drawn normal to these velocities gives us the instantaneous centre IC3.

Graphically, we can find that IC3 is located 236 mm vertically above slider 4.

Velocity V23 is found from the length of member 2 and its angular speed, as

m/s. 6.283mm 60602rpm 1000

mm 6023

=

=

=

V

Thus the angular speed of member 3 can be found (the distance of pin 23 to IC3 can be measured to be approximately 210 mm):

rad/s. 9.29m 210.0/m/s 6.283

/233

==

= rV

Then velocity V4 is

right. m/s 7.061m 236.0rad/s 29.9 mm 23634

==

= V

- 2 -

Question 2

To solve this problem, we will use a velocity diagram to determine the velocity of pin 34 with respect to ground, V1,34, and then use a distance ratio between the blade tip and pin 34, and the point on member 4 about which they both rotate, to determine the blade tip velocity.

Firstly, V1,23 can be determined from given information in the question:

mm/s. 3.340mm 65602rpm 50

mm 6523,1

=

=

=

V

Then V1,34 can be measured from the diagram:

mm/s. 19934,1 =V

Then the blade tip velocity can be found:

mm/s. 613 08.334,14 ==VV

- 3 -

Question 3

As with Questions 1, the locations of the instantaneous centres of members 2 and 5 need to be found. Member 2 can be solved first, as velocity directions at pin 2-3 and slider 4 can be determined immediately. The instantaneous centre IC2 is located below and to the left of slider 6.

Once IC2 is known, the direction of velocity of pin 2-5 can be found (V25), and once this is known, the instantaneous centre of member 5 can be determined.

IC5 is located slightly above slider 6.

- 4 -

Taking measurements from the images allows us to determine the required linear and angular speeds:

Member 2 angular speed:

.rad/s 39.16m 305.0/m/s 5

/42

=

=

= rV

Member 3 angular speed:

.rad/s 19.39m 138.0/m 33.0

m 138.0//

2

23

233

=

=

=

=

V

rV

Member 5 angular speed:

.rad/s 58.36m 138.0/m 308.0

/

2

255

=

=

=

rV

Slider 6 speed:

.m/s 38.2 mm 6556

=

= V

- 5 -

Question 4 First we will calculate the velocity of point C:

m/s. 0.0335

m 32.0602rpm 1

m 32.0

=

=

=

CV

The velocity at point B can be determined by constructing an appropriate velocity diagram (note that the direction of velocity of point C w.r.t. point B is being used, and recognise that the direction of the velocity at point B is known as member 4 is rotating about the ground):

Measuring the velocities gives:

m/s. 0534.0m/s 0335.0595.1

=

=BV

Now we can draw lines in the directions of velocity of point N w.r.t. both B and C from the ends of the arrows in the above diagram, to find the velocity of point N w.r.t. ground:

- 6 -

Now we can measure the velocity at point N:

m/s. 06.0m/s 0335.079.1

=

=NV

- 7 -

The following are solutions to the supplementary questions.

Question 5

To solve this problem using Velocity Diagrams, we need to find V1,4 (the velocity of member 4 w.r.t. member 1 (ground).

Mathematically, this vector can be obtained by the sum

V1,4 = V1,23 + V23,4.

We know the direction of V14, (horizontal line) and the velocity V1,23 can be ascertained immediately from the information given in the question as

m/s. 6.283mm 60602rpm 1000

mm 6023,1

=

=

=

V

From the origin (representing the ground - 1), draw a vector for V1,23, and a horizontal line for the direction of V1,4:

For member 3, we do not know, in advance, the relative velocity of the pin connecting members 3 & 4 with respect to the pin connecting members 2 & 3. We do know its direction (normal to the line connecting the two pins, as member 3 is a rigid member, and pin 34 is rotating around pin 23). Due to the vector sum, this velocity component should pass through the head of vector V1,23:

1

2 3

4

1

2 3

4

- 8 -

Finally we can determine V1,4:

By taking measurements from this diagram, the magnitude of V1,4 can be determined:

m/s. 6.974,1 =V

- 9 -

Question 6 (i) To find the speed of the tip of the blade, we need to determine the following:

The distance of the blade tip from its instantaneous centre (the pin connecting member 4 to the ground)

o This distance can be measured from the diagram as approximately 370 mm. The velocity of pin 34 (V34)

o This requires knowledge of The angular velocity (3) and instantaneous centre (IC3) of member 3

o To find 3 we need the velocity of pin 23 (V23), and the distance of pin 23 from IC3. o IC3 can be determined by finding the intersection of lines normal to the velocities of pins 23

and 34 (which are both points on member 3). Graphically, this point lies in line with both members 2 and 4, and is located a little above the blade tip.

V23 can be determined from given information:

mm/s. 3.340mm 65602rpm 50

mm 6523

=

=

=

V

Now, pin 23 is located approximately 545 mm from IC3 (measured from the diagram), which allows us

to determine 3:

rad/s. 6244.0mm 545/mm/s 3.340

/233

==

= rV

Given 3, and the measured distance from IC3 to pin 34 (310 mm), we can now determine the velocity

at pin 34:

mm/s, 6.193mm 310rad/s 0.6244 mm 310334

==

=V

and then without explicitly evaluating 4, we can determine the blade tip velocity from the ratio of radial distances from the instantaneous centre of member 4:

mm/s. 597 mm 120/mm 370344

=

=VV

(ii) To find the angle through which the blade would rotate (graphically), we must first appreciate under what arrangements the blade will be at its nearest and farthest extents.

These will occur when members 2 and 3 are parallel, with maximum extent occurring when member 2 is pointing upwards and to the right, and the members are end-to-end, and minimum extent occurring when member 2 pointing down and left, and member 3 is overlapping member 2.

Under these conditions, the effective combined length of member 23 would be L3+L2, or L3-L2, respectively. From the diagram, these can be measured as L2 = 65 and L3 = 288, thus L3+L2 = 353 and L3-L2 = 223. As member 3 is connected via pin 34 to member 4, the task is to find at what angles member 4 would be when pin 34 is these distances from the pin about which member 2 is rotating.

We can make an estimate graphically, but a more precise indication can be gained geometrically by constructing triangles, and solving for the unknown angles:

- 10 -

Minimum angle case Maximum angle case

To solve for the angles, use the Law of Cosines (note: this formula should be provided in the Practice Classes):

( )cos2222 caacb +=

Thus:

( )( )

o

o

2.113,

5.4681086.0

68887442.0coscos1202882120288223

2

1

1

1222

=

==

=

+=

Likewise

Therefore the angle through which the blade rotates is

113.2 - 46.5 = 66.7.

(iii) By inspection, it can be seen that the angle would increase if the length ratio between members 2 and 4 was increased.

- 11 -

Question 7 We will use a velocity diagram to determine velocity of pin 23, by realising that we know the velocity of slider 4 w.r.t. ground, and know the direction of the relative velocity of pin 23 w.r.t. pin 24.

From this diagram, we can measure V1,23 as:

m/s. 4.97 993.0m/s 523,1

=

=V

How do we find the velocity of pin 25? We realise that this pin lies on a line on member 2 connecting pins 24 and 23. Pin 25 rotates around pin 24 just as pin 23 does, but it is closer than pin 23, and thus by taking a ratio of distances, we can determine the magnitude of this relative velocity:

Measuring from the diagram gives the velocity of pin 25:

m/s. 4.67 935.0m/s 525,1

=

=V

- 12 -

Finally, a velocity diagram can also be constructed to determine the velocity of slider 6 (note that we know that slider 6 is constrained to move horizontally w.r.t. ground): Once IC2 is known, the direction of velocity of pin 2-5 can be found (V25), and once this is known, the instantaneous centre of member 5 can be determined.

The slider velocity can be measured from the diagram to be:

m/s. 4.16,1 =V

- 13 -

Question 8 Find the directions of velocities points B and C, and use these to determine the location of the instantaneous centre for member 3: We can also use this instantaneous centre to determine the velocity at point N: Taking measurements from the images allows us to determine the required linear and angular speeds:

Member 3 angular speed:

.rad/s 0838.0m 4.0/

/m/s 0.0335

m 32.0602rpm 1

m 32.0

3

=

=

==

=

=

C

C

C

VrV

V

Speeds at B and N:

.m/s 0.0693m 827.0

m/s 0.0525m 627.0

=

==

=

N

B

V

V

Member 4 angular speed:

.rad/s 158.0m 333.0/

/4

=

=

=

B

B

VrV