Example using equation 1
• An animal jumps vertically. The animal’s takeoff
or initial velocity is 3 m/s. How long does it take
to reach the highest point of the jump? (Air
resistance negligible).
A) - 0.3 s
B) 0.03 s
C) - 0.03 s
D) 0.3 s
Example using equation 1
• An animal jumps vertically. The animal’s takeoff
or initial velocity is 3 m/s. How long does it take
to reach the highest point of the jump? (Air
resistance negligible).
• vi = 3 m / s; a = -9.81 m / s2
• At point when animal is highest: vf = 0
• vf = vi + at ---> t = (vf - vi) / a
• t = (0 - 3) / (-9.81)
= 0.3 seconds
Example using equation 2
What was the net jump height of the animal in
example #1. In other words, how high did it jump
once its feet left the ground?
A)
B)
C)
D)
46 m
46 cm
1.34 m
-1.34 m
Example using equation 2
• What was the net jump height of the animal in
example #1
Knowns: vi = 3 m/s; t = 0.3 sec, a = -9.81 m/s2
Want to solve for rf
rf - ri = vit + 1/2 * at2
ri = 0 when legs are extended
rf = 3t + (0.5) (-9.81) (t2)
rf = 3 *(0.3) + (0.5) (-9.81) (0.3)2
rf = 0.9 + (0.5) (-9.81) (0.09)
rf = 0.9 + (-0.44) = 0.46 m about 18 inches
Example using Equation 3
A runner starts at a standstill and accelerates
uniformly at 1 m/s2. How far has she travelled at
the point when she reaches her maximum speed
of 10 m/s?
A) 5 m
B) 10 m
C) 50 m
D) 100 m
Example using Equation 3
• A runner starts at a standstill and accelerates
uniformly at 1 m/s2. How far has she travelled
at the point when she reaches her maximum
speed of 10 m/s?
• Given: vmax= 10 m/s; a=1 m/s2; Dr=?
vf2 = vi2 + 2a (rf - ri)
∆r = (vf2 - vi2)/ 2a
∆r = (102 - 02) / (2 * 1)
∆r = 100 / 2 = 50 meters
A person sits in a chair and does a knee extension. The
knee angle changes by 0.5 radians in 0.5 seconds. What
is the magnitude of the angular velocity of the knee, the
linear velocity of the foot and the linear displacement of
the foot? (Shank length is 0.5 meters)
A) 1 rad/s, 0.5 m/s, 0.25rad
A) 0.5 rad/s, 0.5 rad/s, 0.25rad
A) 1 rad/s, 0.5 m/s, 0.25 m
A) 0.25 rad/s, 0.5 m/s, 0.25m
• Given:
∆ =0.5rad
∆t = 0.5s
r=0.5m
• Angular velocity of the knee
 = ∆ / ∆t = 0.5 rad / 0.5 s = 1 rad / s
• Linear velocity of the foot
 v = r = (1 rad/s) * (0.5 m) = 0.5 m/s
• Linear displacement of foot
 s = r = (0.5 meters)(0.5 rad) = 0.25 m
A 100 kg person is riding a bike around a corner (radius = 2 m)
with a constant angular velocity of 1 rad /sec. What were the
magnitudes of the tangential & radial components of the
acceleration?
A) at = 981 m/s2; ar = 2 m/s2
A) at = 0 m/s2; ar = 4 m/s2
A) at = 0 m/s2; ar = 2 m/s2
A) at = 0 m/s2; ar = 2 rad/s2
E) at = 981 m/s2; ar = 4 rad/s2
A 100 kg person is riding a bike around a corner (radius = 2 m)
with a constant angular velocity of 1 rad /sec. What were the
magnitudes of the tangential & radial components of the
acceleration?
Given:
 = 0; r = 2 m;  = 1 rad/sec
at = r
at = r= (2 m)(0 rad/s2) = 0 m/s2
ar = r2
ar = (2 meters) (1 rad/s)2 = 2 m/s2
Angular Kinematics Example # 1
A 60 kg sprinter is running a race on an unbanked circular track with a circumference
of 200 m. At the start, she increases speed at a constant rate of 5 m/s2 for 2 seconds.
After two seconds, she maintains a constant speed for the duration of the race.
a. How long does it take for her to run the 200 m?
b. What is her speed when she crosses the finish line?
c. What is her angular velocity when she finishes?
d. What was her angular acceleration during the first two seconds?
e. What is her angular acceleration when she finishes?
f. Determine her radial and tangential accelerations 1 second after the start of the
race.
Angular Kinematics Example # 1
A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she increases speed at a constant rate of 5 m/s2 for 2 seconds.
After two seconds, she maintains a constant speed for the duration of the race.
a.
How long does it take for her to run the 200 m?
Given:
Dr = 200m
m=60kg
From t = 0s to t= 2s at=5 m/s2
From t = 2s to end at=0 m/s2
Solve for total time of race
To solve for the time, you need to break the race into two time periods:
one with a constant tangential acceleration (at) of 5 m/s2 (the first two seconds) and
one with a constant tangential acceleration of 0 m/s2 (after two seconds have gone by).
Find the distance traveled in the first 2 seconds:
Dr = vit + 0.5 at2 = (0)(2) + 0.5(5)(2)2 = 10 m
After the first two seconds, the runner has (200 m - 10 m) = 190 m left to go.
Find the tangential velocity after the 2 seconds of acceleration:
v2sec = v0sec + at = 0 + (5)(2) = 10 m/s
Find the time to complete the remaining distance at this velocity:
Dr = vit + 0.5 at2 = (10)t + 0 = 190 m
t = 19 s
The total time is 19 s plus the first 2 s which equals 21s.
Angular Kinematics Example # 1
A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she
increases speed at a constant rate of 5 m/s2 for 2 seconds. After two seconds, she maintains a constant speed
for the duration of the race.
Given:
Dr = 200m
m=60kg
From t = 0s to t= 2s at=5 m/s2
From t = 2s to end at=0 m/s2
b. What is her speed when she crosses the finish line?
Since there is no tangential acceleration after the first two seconds, her speed is constant. Therefore her speed
when she crosses the finish line is 10 m/s.
c. What is her angular velocity when she finishes?
Since the circumference of the track is 200 m = 2pr , then r = 31.83 m.
Her angular velocity when she finishes is  = v / r = (10m/s) / (31.83m) = 0.31 rad/s
Angular Kinematics Example # 1
A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she
increases speed at a constant rate of 5 m/s2 for 2 seconds. After two seconds, she maintains a constant speed
for the duration of the race.
d. What was her angular acceleration during the first two seconds?
Her angular acceleration during the first two seconds is
 = at / r = (5 m/s2) / (31.83 m) = 0.16 rad/s2
e. What is her angular acceleration when she finishes?
= at / r = (0 m/s2) / (31.83 m) = 0 rad/s2
f. Determine her radial and tangential accelerations 1 second after the start of the race.
To find the radial acceleration after one second, you need to find the angular velocity after one second.
Remember the angular acceleration  = 0.16 rad/s2 (from part d)
and the initial angular velocity 0s = 0
a = D/t
D = 1s – 0s = t
1s = t + 0s = 0.16(1) + 0 = 0.16 rad/s
Therefore her radial acceleration after one second is
ar = r1s2 = (31.83)(.16)2 = 0.81 m/s2
The tangential acceleration after one second is given, at = 5 m/s2
Usain Bolt
• http://www.youtube.com/watch?v=o-urnlaJpOA
How fast could Usain Bolt have run?
A Dynamical Study
• http://www.youtube.com/watch?v=o-urnlaJpOA
• Eriksen et al. 2008 – could 9.69s have been 9.52s?
Could Usain Bolt have run a faster 100m?
• http://www.youtube.com/watch?v=o-urnlaJpOA
• Eriksen et al. 2008 – could 9.62s have been 9.52s?
Could Usain Bolt have run a faster 100m?
• http://www.youtube.com/watch?v=o-urnlaJpOA
• Eriksen et al. 2008 – could 9.69s have been 9.52s?
• Close – but more like 9.55s