1. A long jumper's take-off velocity is 10 m/s at an angle of 20 o
to the horizontal. His centre of gravity is 1 m above the ground at take-off and is 0.5 m above the ground at landing. a) How far does he displace his centre of gravity horizontally between take-off and landing?
[4] b) Is this the optimal angle of take-off? If not, what would the approximate optimal angle be
(no calculation necessary)? Given that this velocity vector is typical of world class long jumpers, why do they not increase their angle of take-off? [4]
2. For this question refer to the diagram and table
(co-ordinate data for the three joint centres of
Time Shoulder (S)
0.00 x y
251.3 220.1
Elbow (E) x
275.8 y
Wrist (W) x y
225.0 262.3 243.6 rotations). Not all this information will be
0.01 251.9 220.1 276.4 225.0 267.8 253.7
0.02 252.5 220.1 277.0 225.0 277.0 255.0 needed in this question. 0.03 253.1 220.1 277.6 225.0 288.1 253.1
0.04 253.7 220.1 278.2 225.0 294.2 250.4
The co-ordinate data is in cm . The forearm segment is 0.3 m in length, which you could have calculated from the data if not supplied. a) What is the angular velocity of the forearm segment at 0.02 seconds? [10] b) If the agonists for this movement in part a) were contracting concentrically which muscles would be the agonists? [2]] c) What is the linear velocity of the wrist joint centre of rotation relative to a fixed frame of reference at time = 0.02 (not relative to the elbow joint)?
[5 d) Calculate the linear acceleration of the shoulder at time = 0.02 seconds? [5]
1

e) If the athlete took off for this jump shot with a centre of gravity velocity of 3 m/s at 80 o
to the right horizontal, how high will he jump and how far would he travel horizontally
(assume the C of g height is the same at take-off and landing.
[8]
3. A person throws a discus (mass = 2 kg) at an angle of 40 o
to the horizontal with a velocity of 20 m/s. Calculate the distance the object travelled horizontally prior to ground contact
(neglecting air resistance) if the discus was released 1.5 m above the ground. [7]
4. List four possible errors we must account for when analysing human motion using film, which will later be digitised to give joint centre displacement (co-ordinate data) and then that data will be differentiated to obtain velocity and acceleration. [4]
5. Why do we need to filter kinematic data such as positional co-ordinate data of joint centres of rotation? [4]
6. The following question was presented as an assignment due to its length.
Given co-ordinate data from anatomical markers at either end of a limb segment it is possible to calculate the absolute angle of that segment in space. Given the absolute angles of two adjoining segments it is possible to calculate the relative angle at the joint between those segments. It is not necessary that the two markers be at the extreme ends of the limb segment, as long as they are in line with the segments longitudinal axis.
The figure opposite shows the outline of a leg with seven anatomical markers in a four segment three joint
1 system (note the foot is modelled as two segments).
The following angles can be determined.
Thigh angle (absolute) =
θ
21
Shank angle (absolute) =
θ
43
Foot angle (absolute) =
θ
65
Metatarsal angle (absolute) =
θ
Knee angle (relative) =
θ
21
-
θ
43
76
(+ve for flexion, -ve for extension)
Ankle angle (relative) =
θ
43
-
θ
65
+ 90 o
(+ve for plantarflexion, -ve for dorsiflexion)
Metatarsal-phalangeal angle (relative) =
θ
65
-
θ
76
2
3
θ
21
4
θ
43
7
On the next page is a table of filtered co-ordinate data from a video analysis of a subject walking from left to
5
6 right as indicated in the above diagram. The columns respond to the seven markers shown above (greater trochanter = right hip = marker 1). The video camera frame speed is 60 frames per second.
Basic mathematics will allow you to calculate any of the above angles from the co-ordinate data. For example, the absolute angle of the shank for any given frame is by the equation of the left:
43
= tan
−
1 y
3
− y
4
−
4
i
=
i
+
1
−
2
∆ t
i
−
1
i
=
2
∆ t i
−
1
2

All marker positions (x, y) are in cm greater trochanter femoral condyle tibial condyle lateral tarsal malleolus calcaneous /metatarsal toe
FRAME TIME [s] x y x y x y x y x y x y x y
-1.0 0.0000 120.2 93.4 109.8 54.3 106.0 51.0 88.8 15.3 79.8 8.2 101.0 0.9 106.8 2.2
0.0 0.0167 122.8 93.1 112.2 54.2 108.3 50.9 89.7 16.0 80.3 9.4 100.9 0.9 106.8 2.2
1.0 0.0333 125.4 92.9 114.9 53.9 111.0 50.8 90.8 16.8 81.1 10.9 100.9 0.9 106.7 2.2
2.0 0.0500 128.1 92.6 118.0 53.6 114.0 50.6 92.1 17.6 82.1 12.6 100.8 1.0 106.6 2.1
3.0 0.0667 130.9 92.2 121.3 53.2 117.4 50.3 93.7 18.6 83.4 14.6 100.7 1.1 106.5 2.0
4.0 0.0833 133.7 91.9 125.0 52.8 121.0 50.0 95.6 19.7 85.1 16.8 100.8 1.4 106.4 1.9
5.0 0.1000 136.6 91.7 129.0 52.4 125.0 49.7 97.9 20.9 87.2 19.1 101.1 1.8 106.6 1.8
6.0 0.1167 139.4 91.4 133.2 52.0 129.3 49.4 100.6 22.1 89.7 21.3 101.8 2.4 107.2 1.8
7.0 0.1333 142.3 91.3 137.6 51.7 133.7 49.2 103.7 23.4 92.6 23.5 103.2 3.2 108.3 1.8
8.0 0.1500 145.1 91.2 142.2 51.5 138.3 49.1 107.1 24.6 96.0 25.3 105.4 4.0 110.2 1.9
9.0 0.1667 147.8 91.2 146.8 51.5 143.0 49.0 110.9 25.6 99.7 26.6 108.5 4.8 113.1 2.1
10.0 0.1833 150.4 91.4 151.1 51.6 147.6 49.2 115.1 26.3 103.8 27.4 112.5 5.4 117.0 2.3
11.0 0.2000 153.0 91.6 155.8 51.9 152.2 49.4 119.5 26.8 108.1 27.5 117.5 5.8 121.8 2.6
12.0 0.2167 155.5 92.0 160.2 52.4 156.6 49.8 124.2 26.8 112.8 27.0 123.2 5.9 127.5 2.7
13.0 0.2333 158.0 92.4 164.4 53.1 160.9 50.4 129.1 26.5 117.7 25.9 129.5 5.8 134.0 2.8
14.0 0.2500 160.5 92.9 168.4 53.8 165.1 50.9 134.2 25.8 122.9 24.3 136.5 5.5 141.0 2.7
15.0 0.2667 162.9 93.4 172.3 54.5 169.0 51.6 139.6 24.7 128.5 22.2 143.5 5.0 148.4 2.6
16.0 0.2833 165.4 93.9 175.9 55.2 172.9 52.2 145.2 23.4 134.3 19.9 151.0 4.3 156.1 2.4
17.0 0.3000 167.8 94.3 179.4 55.9 176.5 52.7 151.0 22.0 140.5 17.3 158.7 3.7 164.0 2.2
18.0 0.3167 170.1 94.7 182.8 56.5 180.0 53.2 157.0 20.4 146.9 14.7 166.5 3.0 171.9 2.1
19.0 0.3333 172.4 94.9 185.9 57.0 183.3 53.6 163.2 18.8 153.6 12.1 174.3 2.5 179.8 2.0
20.0 0.3500 174.7 95.1 188.9 57.4 186.4 53.9 169.6 17.3 160.5 9.8 182.0 2.1 187.7 2.2
21.0 0.3667 177.0 95.2 191.7 57.6 189.4 54.0 176.0 16.0 167.5 7.7 189.7 1.9 195.5 2.4
22.0 0.3833 179.2 95.2 194.4 57.7 192.2 53.9 182.5 14.9 174.7 5.9 197.3 1.9 203.1 2.9
23.0 0.4000 181.3 95.2 196.9 57.7 194.9 53.8 188.9 14.0 181.8 4.5 204.7 2.2 210.4 3.6
24.0 0.4167 183.5 95.0 199.3 57.6 197.5 53.5 195.3 13.3 188.9 3.5 211.9 2.7 217.5 4.5
25.0 0.4333 185.6 94.8 201.5 57.3 200.0 53.1 201.4 13.0 195.8 2.9 218.8 3.5 224.3 5.7
26.0 0.4500 187.8 94.6 203.7 57.0 202.4 52.7 207.3 12.8 202.4 2.5 225.3 4.5 230.6 7.1
27.0 0.4667 190.0 94.2 205.9 56.7 204.8 52.2 212.7 12.9 208.6 2.3 231.3 5.7 236.4 8.6
28.0 0.4833 192.2 93.8 208.1 56.3 207.2 51.7 217.8 13.0 214.3 2.2 236.9 7.1 241.6 10.3
29.0 0.5000 194.5 93.5 210.4 55.9 209.7 51.2 222.3 13.2 219.5 2.2 241.8 8.3 246.3 11.9
30.0 0.5167 197.0 93.1 212.8 55.5 212.2 50.8 226.3 13.4 223.9 2.2 246.0 9.4 250.3 13.2
31.0 0.5333 199.5 92.7 215.2 55.2 214.8 50.5 229.7 13.4 227.6 2.2 249.5 10.1 253.6 14.1
32.0 0.5500 202.1 92.4 217.9 55.0 217.5 50.2 232.6 13.3 230.6 2.1 252.4 10.3 256.3 14.5
33.0 0.5667 204.8 92.1 220.6 54.8 220.2 49.9 234.9 13.1 232.8 2.0 254.6 10.1 258.5 14.2
Left heel strike occurs at frame 1.
Right toe-off occurs at frame 10.
Right heel strike occurs at frame 34.
3

For this question calculate velocity and acceleration using the equations above (first central difference method). This way you can calculate velocities and accelerations that occur at the same time, as was discussed in lecture. a) Calculate the right knee angle at frame 14. Would you describe the knee as flexed or extended? In relation to the gait cycle explain why your answer makes sense. [5] b) Calculate shank angular velocity at frame 24. Again explain this result in relation to the gait cycle.
[5] c) Calculate ankle angular acceleration at frame 7. Discuss your results. Is the ankle plantar flexing or dorsiflexing? Be careful. [10]
7. A cyclist is negotiating a curve with radius 15m at a constant linear speed of 5 m/s. a) Acceleration is occurring. Is it linear or angular acceleration? [1] b) What is the value of this acceleration and in which direction does it occur? [4] c) What is an important factor in determining whether there is enough force to accelerate the bike in this manner: the coefficient of rolling or sliding friction? Explain your choice.
Note that this is really a Kinetics question. [2]
8. Define absolute and relative angles with reference to the human body. Give an example where each of these ways of defining an angle may be useful.
9. a) The table opposite gives the displacement-time Time (s) data for a 100 m sprint race. Assume the athlete 0.0 starts at time 0.0 seconds with zero velocity.
Draw graphs of time-displacement, velocity-time and acceleration-time.
[10]
0.5
1.0
1.5
2.0 b) If the athlete weighed 70 kg what would the forcevelocity graph of this activity look like? Why is this graph not the same shape as the force-velocity
2.5
3.0
3.5 graphs we saw in our look at muscle mechanics? [4]
Note: Due to the large
∆ t in the data opposite, do not use the first central differences method. Instead use a simple finite differences method (example
4.0
4.5
5.0
5.5 below). This will mean you have velocities calculated for different times that the displacement
6.0
6.5 and acceleration data . 7.0
7.5
8.0
V i
+
0 .
5
= d i
+
1
∆
− t d i
8.5
9.0
9.5
Displ. (m)
0.000
0.857
3.160
6.484
10.564
15.210
20.266
25.603
31.130
36.774
42.480
48.213
53.951
59.688
65.422
71.160
76.897
82.635
88.373
94.111
10.0
10.5
99.844
105.514
4